(i - j %k + k) % k 与 (i - j + k )% k

代码

#include<bits/stdc++.h>
using namespace std;

inline int read() {
    int ans = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-')f = -1;
        ch = getchar();
    }
    while (ch <= '9' && ch >= '0') {
        ans = ans * 10 + ch - '0';
        ch = getchar();
    }
    return ans * f;
}
int main() {
    int k = 200;
    for (int i = 1; i<= 10000; i++) {
        for (int j = 1; j<= 10000; j++) {
            if ((i - j + k) %k != ((i-j)%k+k)%k) {
                cout<<i<<" "<<j<<" "<<(i - j + k) %k<<" "<<((i-j)%k+k)%k<<"\n";
            }
        }
    }
    return 0;
}

虽然看起来差不多 , 但是走一步加一个mod运算更加保险