BledDest's blog

By BledDest, history, 4 months ago, translation, In English,
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
Tutorial is loading...
 
 
 
 
  • Vote: I like it  
  • +24
  • Vote: I do not like it  

»
4 months ago, # |
  Vote: I like it -26 Vote: I do not like it

Is the round rated?

»
4 months ago, # |
Rev. 2   Vote: I like it +3 Vote: I do not like it

Another nice idea for solving D can be — link .

»
4 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Can someone tell me why I get wrong answer here in problem D?

http://codeforces.com/contest/846/submission/30118738

  • »
    »
    4 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Sorry, the size of the array wasn't enough.

»
4 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone explain B more explicitly?

  • »
    »
    4 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Suppose you know how many tasks you are solving completely, say i. This costs you i times the sum of costs over all subtasks, and gives you i*(k+1) points.

    For the remaining time, you are not solving complete tasks. How can you do separate subtasks most efficiently? They all give 1 point, so greedily keep solving the lowest time cost one until you have not enough time left.

    So try every possible number for i and take the maximum.

    • »
      »
      »
      4 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Why not solve maximum number of tasks completely and then solve the remaining tasks greedily till we run out of time? I tried doing this, but didn't work.

      • »
        »
        »
        »
        4 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        For example input: 10 2 10 1 9

        In this case you can solve 10 times subtask 1 (10 points) which is better than doing maximum number of tasks completely; that would be solving a single task completely (2 + 1 = 3 points).

        • »
          »
          »
          »
          »
          4 months ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          Thank you!

          • »
            »
            »
            »
            »
            »
            3 months ago, # ^ |
              Vote: I like it 0 Vote: I do not like it

            Maybe O(n2) can solve this problem instead of n2k...

            #include<stdio.h>
            #define S(x,y) (x^=y^=x^=y)
            #define go(i,a,b) for(int i=a;i<=b;i++)
            int n,k,a[50];long long Time,sum,Score,ans,m;
            int main()
            {
            	scanf("%d%d%d",&n,&k,&m);
            	go(i,1,k)scanf("%d",a+i),sum+=a[i];
            	go(i,1,k)go(j,i,k)if(a[i]>a[j])S(a[i],a[j]);
            	
            	go(i,0,n){Score=(k+1)*i;if((Time=sum*i)>m)break;
            	go(j,1,k)Time+(n-i)*a[j]>=m?Score+=(m-Time)/a[j],j=k:(Time+=(n-i)*a[j],Score+=n-i);
            		Score>ans?ans=Score:1;}printf("%I64d\n",ans);return 0;
            }//Paul_Guderian
            
            
»
4 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Please elaborate the logic for Problem A.

»
4 months ago, # |
  Vote: I like it +1 Vote: I do not like it

can someone please explain problem C !

»
4 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Problem C can be solved in linear time. Solution O(n).

  • »
    »
    4 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Elaborate it please..

    • »
      »
      »
      4 months ago, # ^ |
      Rev. 2   Vote: I like it 0 Vote: I do not like it

      First let's solve subtask: Given an array of length N, for each 0 <= i < n we need to find k such that sum[0, k) — sum[k, i] = max.

      Assume we store array A, where A[k] = sum[0, k) — sum[k, i]. Answer for current i is the maximum in A. Now we want to update it for i + 1. To do this reduce all A[0 <= j <= i] by x[i + 1], A[i + 1] = abs(sum[0, i + 1]). So we can use dp, because maximum answer for i + 1 is maximum of (ans[i] — x[i + 1]) and abs(sum[0, i + 1]).

      Then iterate with c through array. It divides into two intervals: [0, c), [c, n). You need to find k1 and k2 such that (sum[0, k1) — sum[k1, c)) + (sum[c, k2) — sum[k2, n)) is maximum. We can use previously calculated dp because this two ranges are independent.

      • »
        »
        »
        »
        4 weeks ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        why Answer for current i is the maximum in A

        • »
          »
          »
          »
          »
          4 weeks ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          I meant A[k] = sum[0, k) — sum[k, i]. Now we want to find answer for current i. Ans[i] = max(sum[0, k) — sum[k, i]) <=> Ans[i] = max(A[k]).

»
4 months ago, # |
  Vote: I like it 0 Vote: I do not like it

E's trick in Input constraints! Missed it.

»
2 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Different approach for F. For each unique value ai in the array find the probability that a random choice for l and r will contain at least one ai. To do this assuming we have all the positions of ai stored we can find the probability of the interval not containing ai by summing each of the probabilities of both l and r being in each of the disjoint intervals not containing ai, which is just the length of the interval squared divided by n2 and subtracting this from 1.

Finally by the linearity of expectation the answer is the sum of these probabilities over all distinct values in the array.