題意：

思路：

代碼：

#include <bits/stdc++.h>using namespace std;typedef long long ll;ll pow_mod(ll a, ll n, ll mod) {    ll res = 1;    while (n) {        if (n & 1) res = res * a % mod;        a = a * a % mod;        n >>= 1;    }    return res;}ll dfs(ll x, ll n, ll mod) {    if (n == 1) return x;    ll mid = n / 2, res;    ll tmp = dfs(x, mid, mod);    if (n & 1) res = (tmp * 10 + x) % mod * pow_mod(10, mid, mod) % mod;    else res = (tmp * pow_mod(10, mid, mod)) % mod;    return (res + tmp) % mod;}int main() {    int T, cs = 0;    scanf("%d", &T);    while (T--) {        ll x, m, k, c;        scanf("%I64d%I64d%I64d%I64d", &x, &m, &k, &c);        PRintf("Case #%d:/n", ++cs);        if (dfs(x, m, k) == c) puts("Yes");        else puts("No");    }    return 0;}