约瑟夫问题自认为的最简方法

TXFK

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

int main () {
    int n, m;
    cin >> n >> m;
    bool dd[n];
    memset(dd, 0, sizeof(dd));
    int cnt = 0;
    int na = -1;
    int ca = 0;
    while (cnt != n) {
        na = (na + 1) % n;
        if (dd[na])    continue;
        ca++;
        if (ca == m) {
            dd[na] = 1;
            ca = 0;
            cnt ++;
            cout << na + 1 << ' ';
        }
    }
}