CF1542B. Plus and Multiply

思路

题目就是问能不能找到一个\(x\)使得\(a^x+yb == n\)自己暴力枚举\(x\)即可

ac代码

#include <bits/stdc++.h>

using namespace std;
using i64 = long long;
const i64 inf = 1e18;
typedef pair<int, int> pii;
const int mod = 1e9 + 7;
const int N = 3e5 + 10;

void solve() {
    i64 a, b, n;
    cin >> n >> a >> b;

    if (a == 1) {
        if ((n - 1) % b == 0) cout << "Yes\n";
        else cout << "No\n";
        return;
    }
    
    if (b == 1) {
        cout << "Yes\n";
        return;
    }

    for (int i = 0; ; i ++) {
        i64 t = pow(a, i);
        if (t > n) break;
        if ((n - t) % b == 0) {
            cout << "Yes\n";
            return;
        }
    }

    cout << "No\n";
}   

int main() {
    ios::sync_with_stdio(0); cin.tie(0);
    cout.tie(0);

    int t = 1;
    cin >> t;
    while (t --) solve();

    return 0;
}