E. Increasing by Modulo

题意

一个长度为 n n n 的数组a，每次可以选择一部分 a i a_i ai ，使得 a i = ( a i + 1 ) m o d m a_i = (a_i + 1)\space mod \space m ai=(ai+1) mod m ，求最少需要几次操作使得这个数组单调不减。

题解

二分操作次数；
每个数都有 m i d mid mid 次操作机会，保证单调递增的同时尽量让当前的数尽可能的小。

代码

#include <bits/stdc++.h>
#define rep(i, a, n) for (int i = a; i <= n; ++i)
#define per(i, a, n) for (int i = n; i >= a; --i)
#ifdef LOCAL
#include "Print.h"
#define de(...) W('[', #__VA_ARGS__,"] =", __VA_ARGS__)
#else
#define de(...)
#endif
using namespace std;
typedef long long ll;
const int maxn = 3e5 + 5;
int n, m, a[maxn];
bool judge(int mid) {int mx = 0;rep(i, 1, n) {if (a[i] > mx) {if (mx + m - a[i] > mid) mx = a[i];} else {if (mx - a[i] > mid) return 0;}}return 1;
}
int solve() {scanf("%d%d", &n, &m);rep(i, 1, n) scanf("%d", &a[i]);int l = 0, r = m;while (l < r) {int mid = (l + r) >> 1;if (judge(mid)) r = mid;else l = mid + 1;}printf("%d\n", r);return 0;
}
int main() {#ifdef LOCALfreopen("in.in", "r", stdin);freopen("out.out", "w", stdout);
#endifint T = 1;// scanf("%d", &T);while (T--) solve();return 0;
}

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E. Increasing by Modulo

E. Increasing by Modulo

题意

题解

代码

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