Идея: BledDest
Разбор
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Решение
def f(x):
x += 1
while(x % 10 == 0):
x //= 10
return x
a = set()
n = int(input())
while(not(n in a)):
a.add(n)
n = f(n)
print(len(a))
Идея: BledDest
Разбор
Tutorial is loading...
Решение
#include<bits/stdc++.h>
using namespace std;
int f[10];
string s;
int main()
{
int n;
cin >> n;
cin >> s;
for(int i = 1; i <= 9; i++)
cin >> f[i];
vector<int> diff;
for(int i = 0; i < n; i++)
diff.push_back(f[s[i] - '0'] - (s[i] - '0'));
for(int i = 0; i < n; i++)
if(diff[i] > 0)
{
while(i < n && diff[i] >= 0)
{
s[i] = char(f[s[i] - '0'] + '0');
i++;
}
break;
}
cout << s << endl;
}
1157C1 - Increasing Subsequence (easy version)
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
string res;
int l = 0, r = n - 1;
int lst = 0;
while (l <= r) {
vector<pair<int, char>> cur;
if (lst < a[l]) cur.push_back(make_pair(a[l], 'L'));
if (lst < a[r]) cur.push_back(make_pair(a[r], 'R'));
sort(cur.begin(), cur.end());
if (int(cur.size()) == 2) {
cur.pop_back();
}
if (int(cur.size()) == 1) {
if (cur[0].second == 'L') {
res += 'L';
lst = a[l];
++l;
} else {
res += 'R';
lst = a[r];
--r;
}
} else {
break;
}
}
cout << res.size() << endl << res << endl;
return 0;
}
1157C2 - Increasing Subsequence (hard version)
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
string res;
int l = 0, r = n - 1;
int lst = 0;
while (l <= r) {
vector<pair<int, char>> cur;
if (lst < a[l]) cur.push_back(make_pair(a[l], 'L'));
if (lst < a[r]) cur.push_back(make_pair(a[r], 'R'));
sort(cur.begin(), cur.end());
if (int(cur.size()) == 2 && cur[0].first != cur[1].first) {
cur.pop_back();
}
if (int(cur.size()) == 1) {
if (cur[0].second == 'L') {
res += 'L';
lst = a[l];
++l;
} else {
res += 'R';
lst = a[r];
--r;
}
} else if (int(cur.size()) == 2) {
int cl = 1, cr = 1;
while (l + cl <= r && a[l + cl] > a[l + cl - 1]) ++cl;
while (r - cr >= l && a[r - cr] > a[r - cr + 1]) ++cr;
if (cl > cr) {
res += string(cl, 'L');
} else {
res += string(cr, 'R');
}
break;
} else {
break;
}
}
cout << res.size() << endl << res << endl;
return 0;
}
1157D - N Problems During K Days
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, k;
cin >> n >> k;
if (n < k * 1ll * (k + 1) / 2) {
cout << "NO" << endl;
return 0;
}
int nn = n - k * (k + 1) / 2;
vector<int> a(k);
for (int i = 0; i < k; ++i) {
a[i] = i + 1 + (nn / k) + (i >= k - nn % k);
}
if (nn != k - 1) {
cout << "YES" << endl;
for (int i = 0; i < k; ++i) cout << a[i] << " ";
cout << endl;
} else {
if (k > 3) {
--a[1];
++a[k - 1];
}
if (k == 2 || k == 3) {
cout << "NO" << endl;
} else {
cout << "YES" << endl;
for (int i = 0; i < k; ++i) cout << a[i] << " ";
cout << endl;
}
}
return 0;
}
Идея: vovuh
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
vector<int> a(n);
multiset<int> b;
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
for (int i = 0; i < n; ++i) {
int x;
cin >> x;
b.insert(x);
}
for (int i = 0; i < n; ++i) {
auto it = b.lower_bound(n - a[i]);
if (it == b.end()) it = b.begin();
cout << (a[i] + *it) % n << " ";
b.erase(it);
}
cout << endl;
return 0;
}
1157F - Maximum Balanced Circle
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n;
cin >> n;
vector<int> a(n);
vector<int> cnt(200 * 1000 + 1);
for (int i = 0; i < n; ++i) {
cin >> a[i];
++cnt[a[i]];
}
sort(a.begin(), a.end());
a.resize(unique(a.begin(), a.end()) - a.begin());
int l = 0, r = 0;
int ans = cnt[a[0]];
for (int i = 0; i < int(a.size()); ++i) {
int j = i + 1;
int sum = cnt[a[i]];
while (a[j] - a[j - 1] == 1 && cnt[a[j]] > 1) {
sum += cnt[a[j]];
++j;
}
int cr = j - 1;
if (j < n && a[j] - a[j - 1] == 1) {
sum += cnt[a[j]];
cr = j;
}
if (ans < sum) {
ans = sum;
l = i;
r = cr;
}
i = j - 1;
}
cout << ans << endl;
for (int c = 0; c < cnt[a[l]]; ++c) cout << a[l] << " ";
for (int i = l + 1; i < r; ++i) {
for (int c = 0; c < cnt[a[i]] - 1; ++c) cout << a[i] << " ";
}
for (int c = 0; l != r && c < cnt[a[r]]; ++c) cout << a[r] << " ";
for (int i = r - 1; i > l; --i) cout << a[i] << " ";
cout << endl;
return 0;
}
1157G - Inverse of Rows and Columns
Идея: vovuh
Это комментарий по поводу квадратичного решения. Спасибо за упоминание этого факта, STommydx!
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int n, m;
void invRow(vector<vector<int>> &a, int idx) {
for (int pos = 0; pos < m; ++pos) {
a[idx][pos] ^= 1;
}
}
void invCol(vector<vector<int>> &a, int idx) {
for (int pos = 0; pos < n; ++pos) {
a[pos][idx] ^= 1;
}
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
cin >> n >> m;
vector<vector<int>> a(n, vector<int>(m));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
cin >> a[i][j];
}
}
for (int cnt0 = 0; cnt0 <= m; ++cnt0) {
string r(n, '0');
string c(m, '0');
vector<vector<int>> b = a;
for (int j = 0; j < m; ++j) {
if ((j < cnt0 && b[0][j] == 1) || (j >= cnt0 && b[0][j] == 0)) {
invCol(b, j);
c[j] = '1';
}
}
bool ok = true;
if (cnt0 < m) {
for (int i = 1; i < n; ++i) {
int sum = accumulate(b[i].begin(), b[i].end(), 0);
if (sum != 0 && sum != m) {
ok = false;
} else if (sum == 0) {
invRow(b, i);
r[i] = '1';
}
}
} else {
int idx = -1;
for (int i = 1; i < n; ++i) {
int sum = accumulate(b[i].begin(), b[i].end(), 0);
if (sum != 0 && sum != m) {
if (idx != -1) ok = false;
else idx = i;
}
}
if (idx == -1) {
for (int i = 1; i < n; ++i) {
int sum = accumulate(b[i].begin(), b[i].end(), 0);
if (sum == 0) {
invRow(b, i);
r[i] = '1';
}
}
} else {
for (int i = 1; i < idx; ++i) {
int sum = accumulate(b[i].begin(), b[i].end(), 0);
if (sum == m) {
invRow(b, i);
r[i] = '1';
}
}
if (b[idx][0] == 1) {
invRow(b, idx);
r[idx] = '1';
}
for (int j = 1; j < m; ++j) {
if (b[idx][j] < b[idx][j - 1]) {
ok = false;
}
}
for (int i = idx + 1; i < n; ++i) {
int sum = accumulate(b[i].begin(), b[i].end(), 0);
if (sum == 0) {
invRow(b, i);
r[i] = '1';
}
}
}
}
if (ok) {
cout << "YES" << endl << r << endl << c << endl;
return 0;
}
}
cout << "NO" << endl;
return 0;
}
The editorial for Problem G looks like a little the editorial of Problem F
Thank you for mentioning, I just forgot to fix this place :)
it was a greedy contest :D Good round by the way
My solution problem D after contest(sad): 1)Find minimum first element with binary search 2) let's get maximum suffix, when we can add 1 for all numbers on segment, do it, while exist this is segment 3) end O(nLogn), and write easy
https://codeforces.com/contest/1157/submission/53393728
В этот раз, однако, див3 был сложноватый
ну разве что С2 была сложноватой для С
1157D — N Problems During K Days can be solved simply by putting $$$1,2,\dots,k$$$ initially and then trying to put $$$(n-(k*(k+1)/2))/k$$$ elements in each, and then from reverse order adding as many elements in each position as possible, without considering special cases separately.
You can go through my submission for clear understanding.
Anyone help me in finding the cause of TLE in Problem D in my solution.
"erase" for vector works in linear time of size of the whole vector
This problem also has a solution like D:
https://www.codechef.com/SNCK1B19/problems/MAXPRODU
Does anyone have a solution for problem E which does not involve STL structures, specifically without using multiset,set,map or multimap ?
You can use segment trees.
check this solution 53473452
I did it using DSU 78261765 Incase if you need a explanation reply to this comment or DM me.
Problem D: "Then if nn != k — 1 or k = 1 then this answer is correct."
Any explanation for this ?
For problem 1157c1, what if the input is
7 7 4 5 6 3 2 1
The solution code out puts 5 RRRRL But if I discard 7 and take the subsequence [4 5 6 3 2 1],it's possible to get a strictly increasing subsequence of size 6, right?
When is the next div3?
What's wrong with my solution for problem B? https://codeforces.com/contest/1157/submission/53566680 I am getting wrong answer on test case 7
Try for
5
43111
9 2 3 5 1 1 1 1 1
Answer should be : 53999, get it?
I solved D in a different way. Mine
I'm kind of late, but problem D can be easily (more or less) solved by binary searching the first element, and then binary searching every other element as well, from left to right, so that the minimum possible sum is <= n and the maximum possible sum is >= n. My submission
For the D problem, I just initialized array to 1,2,...,K then added (S — (K * (K + 1)) / 2) / N to every element in the array, and then just distributed (S — (K * (K + 1)) / 2) % N from the behind appropriately. Just wondering why there were so less submissions for this problem?
How to solve problem Minimum Array using segment trees??
I solved with segment tree, so I will try to explain my solution:
First we need to sort B array, to be able to find some index j for each $$$A_i$$$ such that $$$A_i + B_l \lt n$$$ for all $$$l \lt j$$$, and $$$A_i + B_r \ge n$$$ for all $$$r \ge j$$$. We'll have only one index j for each $$$A_i$$$, because $$$max(A) + max(B) < 2n$$$.
After that we can divide B array in two parts and find the best among these greedly, using segment tree here to get minimum, and then updating the used element to not be taken more than once.
Submission
can anyone tell me mathematical proof of E i just wanted to know why finding greater element is optimal than finding lesser value element if n-a[i] is not found
It is because for any $$$k\geq n-a[i]$$$ and any $$$0\leq b < n-a[i]$$$, you will have $$$(a[i]+k) \mod n \leq a[i] + (n-1) \mod n < a[i] \leq a[i]+b < n$$$.
In problem E first I used a sorted vector to store the elements of array b and used lower_bound+erase operations to find the appropriate element for the current a[i]. But it gave TLE on test 16. But when I used multiset instead of vector it passed all the tests. Can anyone explain why vector solution gave TLE?
This is because erase function in vector works in linear time(n), but in set or multiset, the data is already sorted so it works in constant time.o(1).
Got it. Thanks bro.
i think i have better solution for D. We can search maximum value of c that sum of array(c, c + 1, c + 2 .... c + k) <= n; then we can run through this array from the end and greedily add 1 while the problem condition is being met. since we have to add a maximum of k units, we will be able to do this.
why don't solve using two pointer approach in problem C1?