题目:
说是通过了百分之六十几测试用例,int应该是够用的
基本思路是将输入的字符串放进数组进行逐一判断
代码:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char const* argv[])
{
int n, N;
scanf("%d", &N);
for (int i = 0; i < N; i++) {
int cn = 0, an = 0, tn = 0;
scanf("%d", &n);
char* a = (char*)malloc(n + 1);
int ch = getchar();//防止enter键的干扰
for (int j = 0; j < n; j++) {
a[j] = getchar();
if (a[j] == 'c')cn++;
else if (a[j] == 'a')an++;
else if (a[j] == 't')tn++;//计数
}
printf("%d\n", cn * an * tn);//结果计算
free(a);
}
return 0;
}