### chokudai's blog

By chokudai, history, 2 years ago,

We will hold AtCoder Beginner Contest 169.

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

• +85

 » 2 years ago, # |   -25 Editorial in english must also be posted as soon as in japanese.
•  » » 2 years ago, # ^ |   -8 Will it make any difference for you ??
•  » » 2 years ago, # ^ |   +48 I'll post my solutions after the round ends.
•  » » » 2 years ago, # ^ |   0 where will You post It ?
•  » » » » 2 years ago, # ^ |   +9 As a CF Blog.
•  » » » 2 years ago, # ^ |   +4 Atcoder should pay you for this...
•  » » » 2 years ago, # ^ | ← Rev. 2 →   0 In F, i take input of the interval as pairs and sorted the vector of pairs ,then find the median of that interval. Then do the necessary arithmetic as pointed out by Geothermal in here. This approrach fails in few test cases. What's wrong with my approach with F? Can anyone point out what am i missing? Link of my submission
•  » » » » 2 years ago, # ^ |   +14 I believe that for the $n$ even case, you select the wrong two indices to form the median. Use $\frac{n-1}{2}$ and $\frac{n}{2}$, rather than $\frac{n}{2}$ and $\frac{n+1}{2}.$
•  » » » » » 2 years ago, # ^ |   0 Your suggesstion did help to pass some test cases which were not passing earlier, but still few test cases are failing. Can't think of counter test case which could Hack my code.submission
•  » » » » » » 2 years ago, # ^ |   +1 You don't need to sort vector of pairs. Simply sort individually.
•  » » » » » » » 2 years ago, # ^ |   0 I know from the editorial but the problem is that I don't understand why my solution is wrong?
•  » » » » » » » » 2 years ago, # ^ |   +1 Your solution is wrong because you have sorted vector of pairs. Let these be pairs. 1 5 2 3 After sorting either they will remain same or 2 3 1 5 There is no possible way answer will be wrong in both the cases because for finding median array needs to be sorted.
•  » » » » » » » » » 2 years ago, # ^ |   0 Thanks, finally i contructed a test case where my code fails. Test Case5 1 3 2 9 2 8 1 6 3 3 Correct Output=5 My Output=7
 » 2 years ago, # |   +1 I hope it would be a great Contest
 » 2 years ago, # | ← Rev. 3 →   +9 Excited that I will learn something new!Update: Learned when to use double when not to use doubles and how to use doubles!
 » 2 years ago, # |   -80 2nd and the 3rd question seems easy but is really tricky... i dont know on which cases my code is getting wa...can anyone please help
•  » » 2 years ago, # ^ | ← Rev. 2 →   -24 its cheating bro!
•  » » 2 years ago, # ^ |   -8 same problem lol
•  » » » 2 years ago, # ^ |   0 I think I will be feeling pretty silly after seeing the editorial of b and d.
•  » » 2 years ago, # ^ |   +12 Don't ruin the name of India more than already it is.
•  » » » 2 years ago, # ^ |   +2 You don't need to look into name of india.
 » 2 years ago, # | ← Rev. 2 →   -23 I'm so sad, it took me less time to D than it took me to do B and I haven't even done C. Everytime, I miss one of B-D. I'm cursed D:
•  » » 2 years ago, # ^ |   -42 I am missing something in B and getting WA again and again. Any hints please!
•  » » » 2 years ago, # ^ | ← Rev. 2 →   +19 I'd love to help but I think that is not allowed (correct me if I'm wrong, I'm new to CP), will definitely help after the contest ends.
•  » » » » 2 years ago, # ^ |   0 Yes please man now
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2 years ago, # ^ |
-10

so you have to realize anytime the answer overflows out of limit(of data type) or become > 10^18. also if there is even 1 0, ans = 0. Code:

# include <bits/stdc++.h>

using namespace std;

typedef long long ll; typedef unsigned long long ull; typedef unsigned int uint; typedef long double ld;

int main() { ios_base::sync_with_stdio(false); cin.tie(0);

int n;
cin >> n;
ll t;
ld ans = 1;
for (int i = 0; i < n; i++)
{
cin >> t;
if (ans != -1)
ans *= t;
if (ans > 1000000000000000000 || ans < 0)
ans = -1;
if (t == 0)
ans = 0;
}
cout << (ll)ans;
return 0;


}

•  » » » » » » 2 years ago, # ^ |   0 I forgot <0 :(
•  » » » » » » 2 years ago, # ^ |   +4 I had the same code, only difference was that i used long long for 'ans' instead of long double and i got WA on one of the test cases .Changed to long double and got AC.Why?
•  » » » » » » » 2 years ago, # ^ |   0 Because long long data type could be overflowing from the multiplication of say 10^12 and 10^12.
•  » » » » » » 14 months ago, # ^ |   0 I didn't get why should we create separate case for t==0 isn't it taken care while doing ans*=t ?
•  » » » 2 years ago, # ^ |   0 same here bro..
•  » » » 2 years ago, # ^ |   -9 maybe create a function?
•  » » 2 years ago, # ^ |   0 same here bro :(
•  » » 2 years ago, # ^ |   0 I solved problem C, and I don't even understand why it works.
 » 2 years ago, # |   -9 i cry
 » 2 years ago, # |   -9 w a n n a d i e.
 » 2 years ago, # |   0 Will a CE submission be calculated into the overall penalty?I submitted two pieces of codes in the wrong language and will my overall penalty increase by 10 min?
•  » » 2 years ago, # ^ |   0 you can see it in standings in red next to your solutions.
•  » » 2 years ago, # ^ |   0 No.
 » 2 years ago, # |   +6 How to solve F?
•  » » 2 years ago, # ^ | ← Rev. 2 →   +12 For F, we can notice that ans is simply summation of i from 1 to n of (number of subsets of length i with sum s) * (2 ^ (n — i)).So now let dp[x][j] = summation of i from to n of (number of subsets of length n with sum x) * (2 ^ (n — i)) for the prefix 1..jNow, apply normal knapsack to combine answers, but since you're adding length 1 to your previous sums, we will have to divide them 2dp[x][j] = dp[x][j — 1] + (dp[x — a[j]][j — 1] * modinv(2))Answer is dp[s][n]You can even do it in O(n) space without needing the second dimension, like a normal knapsack, moving backward.
•  » » » 2 years ago, # ^ | ← Rev. 2 →   0 Can you please provide the link for your solution madhav_1999
•  » » » » 2 years ago, # ^ |   +3
•  » » » » » 2 years ago, # ^ | ← Rev. 2 →   0 Thanks you BTW you are my friend now.
•  » » » » » » 2 years ago, # ^ |   0 :)
 » 2 years ago, # |   +21
 » 2 years ago, # | ← Rev. 2 →   +3 How to solve E!
•  » » 2 years ago, # ^ |   +10 If n is odd, get the maximum median (sort array of B's and get its median) and the minimum median (sort array of A's and get it's median). The answer is max median — min median +1.If n is even, sort the arrays and do: max median = B[n/2] + B[(n/2)-1], min median = A[n/2] + A[(n/2)-1]. This way we avoid working with not integer numbers. So the answer is still max median — min median +1.
•  » » » 2 years ago, # ^ |   0 Oh god I was trying to use sweep type algorithm and check for each end point, but your answer is much much easier. Could not implement my approach but will try to refine it
•  » » » 2 years ago, # ^ |   0 Thanks for your approach but can you prove it!
•  » » » 2 years ago, # ^ |   0 I am not able to understand why every value between min and max can be attained? Can you explain plz.
•  » » » » 2 years ago, # ^ |   +4 From the minimum median, you just need to icrease the median by 1 to go to the next value (in case of n pair, increase first the biggest one and after the smaller one in the middle). You can do this until reach the maximum median.
•  » » » 2 years ago, # ^ |   0 Could you please give the proof of this approach?
•  » » 2 years ago, # ^ |   0 Thanks to 1358C, idea is very similar. You can observe minimum median will be median of array x and maximum medium will be median of array y.So values in the range median of array x to median of array y will be covered. Take care when size of array is even.Submission
 » 2 years ago, # |   0 In the second part, there are essentially two tricks for the problem 1) Sort the array. If the value at index i is 0, then we cannot return -1(no matter what). 2) Multiplying two extremely large numbers even when the data type is long long int will result in negative values, which will result in wrong answers. So, a*b>=c is equivalent to b>=(c/a)(take care of the border cases) Link to my code https://atcoder.jp/contests/abc169/submissions/13810433
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 https://atcoder.jp/contests/abc169/submissions/13836888I had an error for just one test case? Can you tell me why?
 » 2 years ago, # |   0 Okay, so in the 3rd question, I simply multiplied a and b, and applied the floor function for the product. Link to my code in C++ https://atcoder.jp/contests/abc169/submissions/13814577
•  » » 2 years ago, # ^ |   0 Submitting the same code, giving WA
 » 2 years ago, # |   +3 I just wonder how to solve B,C...
•  » » 2 years ago, # ^ | ← Rev. 2 →   0
•  » » » 2 years ago, # ^ |   0 In problem C, there can be floating-point error.
•  » » » » 2 years ago, # ^ |   0 But how to avoid??
•  » » » » » 2 years ago, # ^ |   +8 I used string for storing $B$. Then calculated $100*B$. link
•  » » » » » » 2 years ago, # ^ |   0 thank you!
•  » » » » » 2 years ago, # ^ |   +19 Use b=b*100+0.01;instead of b*=100;
•  » » » » » » 2 years ago, # ^ |   0 thank you!
•  » » » » » » 12 months ago, # ^ |   0 Adding the extra 0.01 (or even 0.001) works, but how does it work? 351F44
•  » » » » » » » 11 months ago, # ^ |   0 If we convert double to long long, the fractional part would be discarded. So in this kind of problem, if we add small real number and convert it, we can see the same effect as rounding.
•  » » » » 2 years ago, # ^ | ← Rev. 2 →   0 How will there be a floating point error? And how will b=b*100+0.01 save from it?
•  » » » » » 2 years ago, # ^ |   0 Yes, I also need to know the same.
•  » » » » 21 month(s) ago, # ^ |   0 I cannot understand why is a floating-point error occurring for input: 999990000000001 9.99 in this submission? Here, the result will not cross 10^18. I don't know that this is a reason or not. Can you please tell me why such kind of error occurs with floating-point values?
•  » » » 2 years ago, # ^ |   +8 In your solution for problem B, if resu is 10^17 and next number is 99 then it might overflow
•  » » » » 2 years ago, # ^ |   0 Thank you !
•  » » 2 years ago, # ^ | ← Rev. 3 →   +21 This is sarcasm..Right???
•  » » » 2 years ago, # ^ |   +11 NO NO NO!! I really don't solve them!
•  » » » » 2 years ago, # ^ |   +13 You had a bad day don't worry
•  » » » » » 2 years ago, # ^ | ← Rev. 2 →   +40 . So sad :{
•  » » 2 years ago, # ^ |   0 For B, just check 0 case. Then before multiplying with a[i], check if 1e18/a[i] < cur_product. If it is, simply print -1. Otherwise you can safely multiply.
•  » » » 2 years ago, # ^ | ← Rev. 2 →   0 https://atcoder.jp/contests/abc169/submissions/13804989 I did the same... But getting WA in 2 TCs..
•  » » » » 2 years ago, # ^ |   0 Typecast your 1e18 into int before dividing to avoid precision loss.Changing to ans > (int)(1e18)/i should give you AC
•  » » » 2 years ago, # ^ |   0 Thank you !
 » 2 years ago, # |   0 i was getting TLE Second Question
•  » » 2 years ago, # ^ |   0 I'm guessing you may have tried to multiply all the numbers together in Python or a similar language? Since the intermediate products can have $O(N)$ digits, multiplying them takes $O(N)$ time, so this approach takes $O(N^2)$ time in the worst case.
•  » » » 2 years ago, # ^ |   0 Can you tell what's wrong here and why typecasting solution is not accepted in C++? ll a; double b; cin >> a >> b; cout << (long long)(a*b) << "\n"; 
•  » » » » 2 years ago, # ^ |   0 use long double b;
•  » » » » 2 years ago, # ^ | ← Rev. 3 →   0 This will be WA because of precision error. Instead do (a*(b*100))/100 While calculating b*100, make sure you take round(b*100), otherwise things like 23.999999(due to double's precision) will be floored down to 23.
 » 2 years ago, # |   +29 My solutions to all problems are at https://codeforces.com/blog/entry/78195.
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 This is the solution of the second question. void solve() { cin>>n; vectorv(n); for(ll i=0;i>v[i]; sort(v.begin(),v.end()); ll p=1,f=0; for(ll i=0;i=1000000000 and p>100000000){p=1000000000000000001;break;} if(v[i]>1000000000 and p>=1000000000){p=1000000000000000001;break;} p=(p*v[i]); //if(v[i]) if(p>1000000000000000000)break; } if(p<=1000000000000000000) cout<
 » 2 years ago, # | ← Rev. 2 →   0 Can someone please tell why this fails for C? n = input().split() i1 = int(n[0]) i2 = int(float(n[1]) * 100.0) i = i1 * i2 print(i // 100) 
•  » » 2 years ago, # ^ |   0 You need to round it, rather than taking its floor. For example float('2.51') = 2.509999999..., so int(float('2.51')*100) = 250.
•  » » » 2 years ago, # ^ |   0 thank you sir
 » 2 years ago, # | ← Rev. 2 →   0 In C, why is math.trunc(a*b) in python giving WA,but in CPP, this is accepted- ll a; long double b; cin>>a>>b; ll ans; ans=a*b; cout<
 » 2 years ago, # | ← Rev. 2 →   0 nvm, got it
 » 2 years ago, # |   -7 can anybody find mistake in my D solution. #include using namespace std; bool ispower(int x,set s) { int f=0; int p=1; while(x%2==0) { p*=2; x=x/2; f=1; } if(x==1 && s.find(p)==s.end())return true; if(f==1)return false; int t=sqrt(x); for(int i=3;i<=t;i+=2) { int p=1; while(x%i==0) { f=1; x=x/i; p*=i; } if(f==1 && x==1 && s.find(p)==s.end())return true; if(f==1)return false; } return false; } bool isprime(int x) { for(int i=2;i<=sqrt(x);i++)if(x%i==0)return false; return true; } int main() { long long x; cin>>x; long long ans=0; int r=sqrt(x); set s; bool seive[r+1]; for(int i=2;i<=r;i++)seive[i]=true; for(int i=2;i<=r;i++) { if(seive[i]) { for(int j=i;j<=r;j*=i) { if(x%j==0) { x=x/j; ans++; s.insert(j); } } for(int j=2*i;j<=r;j+=i) { seive[j]=false; } } } if(x>=2 && s.find(x)==s.end() && (ispower(x,s) || isprime(x)))ans++; cout<
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 You are not looking for primes greater than √x.Try for 4000000028.
 » 2 years ago, # | ← Rev. 3 →   0 Any care to explain why is this wrong for problem C ? #define int long long void solve() { int a,b; double c; cin >> a; cin >> c; b = (c*100); int res = (a*b); res/=100; cout << res; } It gives WA for random_12.txtEdit : I always define int as long long in my template.
•  » » 2 years ago, # ^ |   +24 I'd guess the issue is double precision. Precision is an issue in general, but in particular, the answer here can be up to $10^{16}$, which is too large for the number of precision bits in a double.
•  » » » 2 years ago, # ^ |   0 But here multiplications are in long long, the only place where precision could affect is assigning "b = c*100".
•  » » » » 2 years ago, # ^ |   +16 Ah, right; my bad--I misread the solution. The issue is precision in that conversion: I changed b = (c*100) to b = (c*100+0.5) and the solution was accepted.
•  » » » » » 2 years ago, # ^ |   0 Thanks for the solution.Any advice on how to avoid these errors in future ?
•  » » » » » » 2 years ago, # ^ |   +17 You need to understand why this error happened. The reason is that if you read a number as a double (lets say 10) the double can be represented as 9.9999... or maybe 10.000...01 (both of which are 10 indeed woth a very small error intrinsic to the floating point representation)) but in c++ if you convert them to an integer it truncates the value automaticaly, so the result could be either 9 or 10. The solution would be doing the comversion as x_int = round(x_double) or x_int = x_double + 0.5.
•  » » » » » » » 2 years ago, # ^ |   0 Thank you very much. This is what I was looking for.Being skeptic, if x_double is 4.50 is there any chance I will get 5 instead 4 using your suggested solution ?
•  » » » » » » » » 2 years ago, # ^ |   +11 Yes, it would be a problem. But you did not quite get how to use what I said. If a number has at most x decimal digits, you may want to multiply it to 10^x and convert to integer, so you do int_x = 10^x*double_x+0.5 (but if double_x has more the x digits this may not work correctly)
•  » » » » » » » » » 2 years ago, # ^ |   0 Now I got it. Thanks for the help.
•  » » » » » » » » » 2 years ago, # ^ |   -8 but why is this a problem as it is already given in the question that the B is a number with two digits after the decimal point,hence it is always fixed that we are not losing any value for the above solution after multiplying by 100. So why is this so???
•  » » » » » » » 2 years ago, # ^ | ← Rev. 3 →   -8 deeper explaination Essentially, there's some natural imprecision in the way decimal datatypes are stored in C++ (and in most other languages). C++ stores decimals, like all numbers, in binary, rather than base 10, and in binary, numbers like 0.1, 1.6, and 3.4 are actually repeating decimals. Because of that, when C++ tries to store that kind of number, it will actually store a value that's very close, but slightly higher or slightly lower.The problem here, though, is that if you get a value slightly lower than, say, 1.01, then when you multiply it by 100, the result will be slightly lower than 101, and will thus get rounded down to 100, giving you an incorrect answer. To deal with this, we add some value (I used 0.5, but it could be much smaller and still work) to ensure that our result will be larger than 101, rather than smaller. i mean to say in easy words there is no like 3.33 that is actually 3.3333333.....so this is making the answer change .Do comment for any queries the above explanation shows why round function works and floor or ciel function don't credits Geothermal
•  » » » » » » » » 2 years ago, # ^ |   0 credits Geothermal
•  » » » » » » 2 years ago, # ^ |   +8 Do not assume double arithmetic is correct. Often it is not, the result is only near to the mathematic correct result.Deal with that or use integers.
•  » » » » » » » 2 years ago, # ^ |   0 I didn't, that's why I converted double to int and then calculated the result.I am yet to find any solution which always works for doubles.
•  » » » » » » » 2 years ago, # ^ |   +8 I disagree with that. You just need to unserstand how big the error can be in the worst case, often the error is smaller than 1, so you can often get the integer part right for sure if the constraints allow that
•  » » » » » » » » 2 years ago, # ^ |   +13 That is the strategy I would call "deal with it".
•  » » 2 years ago, # ^ |   0 Use long double instead of double
•  » » » 2 years ago, # ^ |   0 Why does long double over double changes anything ? considering value of is B < 10, having 2 digits are decimal point.
•  » » » » 2 years ago, # ^ |   +8 I used this formula: answer = floor( a * long double(b) )
•  » » » » » 2 years ago, # ^ |   0 Yes, that works well. I used the similar thing to get AC later. Thanks for the answer.
•  » » 2 years ago, # ^ |   0 Range of input is too big which will have precision loss. Try using long long and long double
•  » » » 2 years ago, # ^ |   0 Why does long double over double changes anything ? considering value of is B < 10, having 2 digits are decimal point.
 » 2 years ago, # |   0 What is the approach for problem E — Count Median ??
 » 2 years ago, # |   0 In the fourth question, I simply used the Sieve of Eratosthenes to get the prime numbers less than 1e6, after that I iterate from 2 to the largest prime less than 1e6, and simply code out for the number of ways the number can be broken down! Just be careful when we multiply two numbers of digits 5 and 6! Link to my code- https://atcoder.jp/contests/abc169/submissions/13855795
»
2 years ago, # |
-24

what is my mistake in my code???

# include

using namespace std;

int multiply(int array[], int N) { int pro = 1; for (int i = 0; i < N; i++) t = t * array[i]; return t; } int main() { int N; cin>>N; int array[N]; for(int i=0; i<N; i++) { cin>>array[i]; } multiply(array,N); if(array == 0) { cout<<0<<endl; } else if( (multiply(array,N)) > (1000*1000*1000*1000*1000*1000)) {

cout<< -1 <<endl;}
else
{
cout<<multiply(array,N)<<endl;
}

return 0;

}

»
2 years ago, # |
-32

what is my mistake in my code for B???

# include

using namespace std;

int multiply(int array[], int N) { int pro = 1; for (int i = 0; i < N; i++) t = t * array[i]; return t; } int main() { int N; cin>>N; int array[N]; for(int i=0; i<N; i++) { cin>>array[i]; } multiply(array,N); if(array == 0) { cout<<0<<endl; } else if( (multiply(array,N)) > (1000*1000*1000*1000*1000*1000)) {

cout<< -1 <<endl;}
else
{
cout<<multiply(array,N)<<endl;
}

return 0;

}

 » 2 years ago, # |   0 problem C solution : https://atcoder.jp/contests/abc169/submissions/13874016
 » 2 years ago, # |   +1 Why this fails for problem B: https://atcoder.jp/contests/abc169/submissions/13804989
•  » » 2 years ago, # ^ |   0 store 1e18 in a variable and then do the rest. I think that will do the job!!
 » 2 years ago, # |   0 How to solve c? I multiplied a*b and took floor(a*b). What's wrong? submission
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 See, Double can store large values but on the cost of significant digits,in Long you will get more significant digits before decimal point but then you can not store digits after the decimal points
•  » » » 2 years ago, # ^ |   0 But I also tried (long)(b*100)*a/100. What's wrong with this?
 » 2 years ago, # |   +1 problem B solution : https://atcoder.jp/contests/abc169/submissions/13824612
 » 2 years ago, # | ← Rev. 2 →   0 Please someone help me out with Problem C ....... My Submission
•  » » 2 years ago, # ^ |   0 Precision issue. Do long long x = round(b*100) and AC
 » 2 years ago, # |   0 i thought C would give precision error if i use long double. i tried almost 6/7 types of approach(wrong) without using floating point. about 3 minutes before the contest would end, i simply tried using long double and output it with type casting to long long and it worked! wow -_-
 » 2 years ago, # |   0 Whats wrong wit Cimport java.math.BigDecimal; import java.util.Scanner;public class Main { public static void main(String args[]) { Scanner sc=new Scanner(System.in); long a=sc.nextLong(); long b=(long)(sc.nextDouble()*100); long ans=(a*b)/100; System.out.println(ans); }
•  » » 2 years ago, # ^ | ← Rev. 2 →   0 My Simple and short Solution for C (Used Bigdecimal Library in JAVA) void solve() throws Exception{ BigDecimal bd1 = new BigDecimal(ns()); BigDecimal bd2 = new BigDecimal(ns()); bd1 = bd1.multiply(bd2); pn(bd1.toString().split("\\.")[0]); } 
•  » » » 2 years ago, # ^ |   0 You can do it this way but still whats wrong in my code.
•  » » » » 2 years ago, # ^ |   0 Test it out where the floating point value is stuff like 1.12, 1.13, etc (if Java handles floating point the same as python those two cases should show you the issue). Floating point error in this line is your issue:long b=(long)(sc.nextDouble()*100);
 » 2 years ago, # | ← Rev. 7 →   +19 I usually get one hit on both Codeforces and Atcoder, but this time seems horrible xD
•  » » 2 years ago, # ^ |   +1 Dude you are really patient!
 » 2 years ago, # |   0 Whats wrong in Dimport java.util.Scanner;public class Main { public static void main(String args[]) { Scanner sc=new Scanner(System.in); long n=sc.nextLong(); long copyn=n; long x=(long)Math.sqrt(n); long copyx=x; long ans=0; for(int i=2;i<=x;i++) { int count=1; while(n%((Math.pow(i,count)))==0) { n= (long) (n/Math.pow(i,count)); count++; ans++; } while(n%(Math.pow(i,count))==0) { n= (long) (n/Math.pow(i,count)); } } if(copyn==1) { System.out.println(0); } else if(n>x) { System.out.println(1+ans); } else { System.out.println(ans); } }}
»
2 years ago, # |
-15

What is the problem with this solution of C?Can someone help?

# include<bits/stdc++.h>

using namespace std; typedef long long ll;

# define mod 1000000007

int main() { ll a; float b1; cin>>a>>b1;
b1=b1*100; ll b=b1; ll s=a*b; s=s/100; printf("%lld\n",s); }

•  » » 2 years ago, # ^ |   0 Maybe precision issues? Try doing b1 = round(b1*100)
•  » » » 2 years ago, # ^ |   0 Well, if the input is 1.890000 then it becomes 189.000000 and then 189 by typecast. There shouldn't be any problem.
•  » » » » 2 years ago, # ^ |   +7 You dont understand how it works. The number may be interpreted as 188.99999999... which is the same as 189.00000 but when you cast to int in c++ the value is truncated, it sort of an undefined behaviour, the value may be casted to 188 or 189, use the round to avoid that.
•  » » » » » 2 years ago, # ^ |   0 Yeah, I got it. That was the problem actually. Thanks a lot.
•  » » 2 years ago, # ^ |   0 awesome one ! me i just define b as long double and took long long c=a*b and then i cout c and it works also as simple as that
 » 2 years ago, # |   0 Can anybody help me why this submission for D fails at 18th testcase? submission
 » 2 years ago, # | ← Rev. 2 →   0 can some one explain to me why my 1st test case not working of B question
•  » » 2 years ago, # ^ |   0 I have a similar problem too: https://atcoder.jp/contests/abc169/submissions/13892048What is more, the first test case (hand_01.txt) on AtCoder's Dropbox looks like this:Input: 2 4294967296 4294967296Output: -1I compiled my program locally using different compilers and there was nothing wrong with my code ://
 » 2 years ago, # |   0
 » 2 years ago, # |   0 Both are identical solutions for problem B of AtCoder Beginner Contest 169. Does anyone knows why this happen?
•  » » 2 years ago, # ^ |   -10 Maybe, someone playing with the testcases :)
•  » » » 2 years ago, # ^ |   +7 Or maybe first solution is sent to Problem A not Problem B?
•  » » » » 2 years ago, # ^ | ← Rev. 3 →   0 Yes, thanks Mucosolvan
 » 2 years ago, # |   0 I think I got a beautiful solution for F: the answer is just the coefficient of x^s in polynominal (2 + x^a1)(2 + x^a2)...(2 + x^an) here '^' is for power. we can just ignore higher order coefficient than x^s.https://atcoder.jp/contests/abc169/submissions/13847100please ignore unremoved code for problem D, E.
 » 2 years ago, # | ← Rev. 2 →   +5 In B:For the statement: if(inp[i] <= 1e18/ans) I got WA. But when i wrote : if(inp[i] <= 1000000000000000000/ans) I got AC. Is there any differences between 1e18 and 1000000000000000000?? Otherwise what is the reason??
•  » » 2 years ago, # ^ |   +1 You should cast 1e18 to long long otherwise the result of the division becomes a float.
•  » » » 2 years ago, # ^ |   +1 Thanks man. That means 1e18 format is double type.
•  » » » » 2 years ago, # ^ |   0 Yup!
 » 2 years ago, # | ← Rev. 3 →   +8 In editorial there is a typo in problem D English version. Wrong:$O(N)$Correct:$O(sqrt(N))$
 » 2 years ago, # |   +5 Can AnyBody please tell how can i reduce the submission time of my solution to the problem F?https://atcoder.jp/contests/abc169/submissions/13920879I have seen people getting AC in less than 100ms while mine is giving AC in 1000 ms.thanks in advance. :).
 » 2 years ago, # |   0 In the editorial of problem F it says:dp[i][j] = When the choices for the first i options is already determined, the number of combination such that the sum of ak for each k that the first option was chosen is equal to jWhat does it even mean?
•  » » 2 years ago, # ^ |   0 I think it just means, dp[i][j] refers to how many ways we can reach sum equal to j using i elements
 » 2 years ago, # | ← Rev. 2 →   0 Can anyone help me debug my solution? I get TLE on 6-7 test cases, no idea why. Thank you! #include #include using namespace std; int solve(long long n) { if (n == 1) { return 0; } vector f; for (int i = 2; i * i <= n; i++) { while (n % i == 0) { f.push_back(i); n /= i; } } if (n > 1) f.push_back(n); int ans = 0; int i = 0; while (i < f.size()) { int j = i + 1; while (j < f.size() && f.at(j) == f.at(i)) j++; int count = j - i; i = j; int sum = 0; for (int x = 1; count >= sum + x; x++, ans++) sum += x; } return ans; } int main() { long long n; cin >> n; cout << solve(n) << endl; } 
•  » » 2 years ago, # ^ |   +4 for (int i = 2; i * i <= n; i++) i * i can overflow since n goes upto $10^{12}$ Use long long instead.
•  » » » 2 years ago, # ^ |   +1 T_T I don't know how I missed that. Thanks!
 » 2 years ago, # |   0 what is my wrong with my code in problem-c? https://atcoder.jp/contests/abc169/submissions/13926562
 » 2 years ago, # |   0 For problem D, Maybe this sounds stupid, But can someone please explain why is it optimal to compute for each prime and its powers first, then similarly with next prime and its powers, and so on instead of going from 2 onwards ?
•  » » 2 years ago, # ^ |   0 same thing I was also wondering .
 » 2 years ago, # |   0 Can anyone provide any proof for problem-E for both odd and even case.
•  » » 2 years ago, # ^ |   0 If we increase one value by $1$ and leave the rest as it is, then median can stay the same or increase by $1$ for odd $n$ and stay the same or increase by \frac{1}{2}Unable to parse markup [type=CF_MATHJAX] is even. Now let's set $X_i = A_i$ and increase all of them one by one till we reach $X_i = B_i$. At first median equals median of $A_i$, and after all operations equals median of $B_i$ and at each step it will either stay the same or increase by $1$ in case of odd $n$ and by $\frac 12$ for even $n$ thus will reach all integers for odd $n$ and halfintegers for even $n$ from range $[\text{median}(A_1, \ldots, A_n), \text{median}(B_1, \ldots, B_n)]$.To prove we can't reach any other value we only need to see, that median of an integer sequence is an integer for odd $n$ and a half integer for even $n$ and see that we can't reach a value outside that segment, but this is straightforward conclusion from the fact, that if all values increase or stay the same, then the median can only increase or stay the same as well.
 » 2 years ago, # | ← Rev. 3 →   0 For problem F, it says f(T) is the number of different non-empty subsets satisfying the given condition. Now consider the first sample. It was given that f(T) is 2 ({1,2}, {3}). By the definition, shouldn't f(T) be 3 ({1,2}, {3}, {1,3}) ? Unless there is a translation issue.
 » 2 years ago, # |   0 If anyone is still stuck at D here is a Solution
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2 years ago, # |
0

I am Facing problems in the C . No matter what I submit , I am shown wrong answer in 6/22 cases . I am not understanding where is the error or fault exists in that code .. If anyone knows , please give me the solution . My code is given below ::::

# include<stdio.h>

int main() { long long int a,pro1; double b,pro=1.0; scanf("%lld %lf",&a,&b); b=b*100; pro=(a*b); pro1=(long long int)(pro/100); printf("%lld",pro1); return 0; }

•  » » 2 years ago, # ^ |   0 For big values of $a$ the variable $pro$ has simply not enough bits to store the exact result of $a*b$. In this case the double type truncates bits, which results in pro1 beeing some other value than the mathematecally expected one.
 » 2 years ago, # |   0 Can someone help me where I'm going wrong in problem D, I'm getting hand_22 as the only test case as WA. Here is my code https://atcoder.jp/contests/abc169/submissions/14102528 Thanks
•  » » 2 years ago, # ^ |   0 ull i=1; You have this just before the for loop, but it should be inside it, so that the while loop always starts at 1. Since all the primefactors in mp are independent of each other.
•  » » » 2 years ago, # ^ |   +3 Thanks a lot for reading my code and identifying my mistake.
 » 2 years ago, # |   0 Can anyone tell me why I am getting runtime error in this solution https://atcoder.jp/contests/abc169/submissions/14630442
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2 years ago, # |
0

//**B**

# include<bits/stdc++.h>

using namespace std; typedef long long int ll; typedef long double ld; int main() { int t; cin>>t; if(t==0)return cout<<0,0; long double s = 1; vector vec(t); for(int i=0; i<t; cin>>vec[i++]); sort(vec.begin(),vec.end()); for(auto &it: vec){ if(it==0)return cout<<0,0; if(s>1e18/it)return cout<<-1,0; s *= it; } cout<<(ll)(s)<<'\n'; }