2
1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 2 6 2 10 2 2 5 6 1 0 2 7 0 2 2 7 5 10 6 10 2 10 6 1 9

7
379297
#include<string.h>
#include<stdio.h>
int ans[55];
int c1[100], c2[100];
int main()
{
	int a[27], t, i, j, k;
	scanf("%d", &t);
	while(t --){
		memset(c1, 0, sizeof(c1));      //初始化
		memset(c2, 0, sizeof(c2));
		c1[0]=1;
		for(i = 1; i <= 26; i ++){
			scanf("%d", &a[i]);
		
//			if(a[i] == 0){ //跳过数目为0的
//				continue;
//			}
			for(j = 0; j <= 50; j ++)
			{  
				for(k = 0; j+k <= 50&&k<=a[i]*i; k += i)
				{                                       //这里的k <= a[i]*i, 意思是表示a[i]个价值为ｉ的字母的组成的最大价值
					c2[j+k] += c1[j];
				}
			}
			for(j = 0; j <= 50; j ++)
			{
				c1[j] = c2[j];
				c2[j] = 0;
			}
		}
		int ans= 0;
		for(i = 1; i <= 50; i ++){
			ans+= c1[i];
		}
		printf("%d\n", ans);
	}
	return 0;
}