PAT乙级1091 N-自守数 (15 分)

https://pintia.cn/problem-sets/994805260223102976/problems/1071785664454127616

#include <iostream>
#include <cmath>
using namespace std;
int main(){
	int K, N, M, cnt = 0, tmpk;
	cin >> M;
	for(int i = 0; i < M; i++){
		cin >> K;
		tmpk = K, cnt = 0;
		while(tmpk){
			cnt++;
			tmpk /= 10;
		}
		int j, tmp;
		for(j = 1; j < 10; j++){
			tmp = j*K*K;
			if((tmp % (int)pow((double)10, (double)cnt)) == K)
				break;
		}
		if(j < 10)
			cout << j << " " << tmp << endl;
		else
			cout << "No\n";
	}
	return 0;
}