转换mod为除法

Problem - B - Codeforces

对于最后一句话：“>的个数是bn/m"

因为0<=bi+1-bi<m, 那么找>就是找有多少个点 bi/m 从x到x+1（0->1，1->2类似于这样子的），那么这样子到n时前面就有 bn/m 个这样子的点

#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
typedef long long ll;
const int N = 1e6 + 5;
ll a[N];
int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    ll n; cin >> n;
    for (int i = 0; i < n; i++)cin >> a[i];
    ll n1, m, x; cin >> n1 >> m >> x;
    ll sum1 = x % m, sum = 0;
    for (int i = 0; i < n; i++) {
        sum += a[i] % m;
    }
    sum1 += n1/n*sum;
    for (int i = 0; i < n1 % n; i++) {
        sum1 += a[i] % m;
    }
    cout << n1 - sum1 / m;
    return 0;
}