題意

for (int i = 1; i <= \(\ 10^k\); i++) (k += m) %= n

輸出最後的k

題解

行動n次就會回到起點，快速冪

除錯記錄

快速冪寫炸了

#include <cstdio>

using namespace std;

int n, m, k, x;

int pow(int num, int times, int mo){
    int ans = 1, t = num;
    while (times > 0){
        if (times & 1) (ans *=
 
 t) %= mo;
        (t *= t) %= mo;
        times /= 2;
    }
    return ans;
}

int main(){
    scanf("%d%d%d%d", &n, &m, &k, &x);
    k = pow(10, k, n);
    
    for (int i = 1; i <= k; i++)
        (x += m) %= n;
    printf("%d\n", x);
    
    return 0;
}