C 解法, 执行用时: 2ms, 内存消耗: 368KB, 提交时间: 2021-02-21

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#pragma warning(disable :4996)
int main()
{
	long long int a;
	scanf("%lld", &a);
	long long int *p;
	p = (long long int*)malloc(a*sizeof(long long int));
	for (int j = 0; j < a; j++)
	{
		scanf("%lld", &p[j]);
	}
	for (int k = 0; k < a; k++)
	{
		int flag = 0;
		int flag1 = 0;
		long long int r = p[k];
		while (r != 0)
		{
			flag++;
			long long int u = r % 10;
			if (u == 0)
			{
				flag1++;
			}
			else
			{
				if (p[k] % u == 0)
				{
					flag1++;
				}
				
			}
			r = r / 10;
		}
		if (flag == flag1)
		{
			printf("G\n");
		}
		if (flag1 == 0)
		{
			printf("S\n");
		}
		if ((flag1 != 0) && (flag != flag1))

		{
			printf("H\n");
		}

	}
}

C++14 解法, 执行用时: 2ms, 内存消耗: 368KB, 提交时间: 2019-12-06

#include<cstdio>
  
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        long long num, cur;
        scanf("%lld", &num);
        int count = 0, len = 0;
        cur = num;
        while (cur) {
            long long temp = cur % 10;
            if (temp == 0) {
                ++count;
            } else if (num % temp == 0) {
                ++count;
            }
            cur /= 10;
            ++len;
        }
        if (count == len) {
            printf("G\n");
        } else if (count) {
            printf("H\n");
        } else {
            printf("S\n");
        }
    }
    return 0;
}