一定要理解是最外面一层的括号 /(ㄒoㄒ)/~~ eg1. a(aa(aaA)a 就是4 /(ㄒoㄒ)/~~ eg2. a(aa)aaA)a 还是4 /(ㄒoㄒ)/~~

#include <stdio.h> #include <string.h> int main() { int n; scanf("%d", &n); while (n--) { char ar[1003]; scanf("%s", ar); int len = strlen(ar); int sum = 0; int a = 0, b = 0; for (int i = 0; i < len; i++) { if (ar[i] == '(') { a = i+1; break; } } for(int j=a;j<len;j++){ if(ar[j]==')'){ b=j; } } for (; a < b; a++) { if (ar[a] >= 97 && ar[a] <= 122) { sum++; } } printf("%d\n", sum); } return 0; }

#include<stdio.h> #include<string.h> int main() { int n; scanf("%d", &n); while (n--) { char arr[1010]; int num = 0; scanf("%s", arr); int sz = strlen(arr); int a = 0; int total = 0; for (int i = 0; i < sz; i++) { num = 0; if (arr[i] == '(') { for (int j = i + 1; arr[j] != ')'&&j<sz; j++) { if (arr[j] >= 'a' && arr[i] <= 'z') { num++;//记录每一次遇到()里的数目 } if (arr[j+1] == ')') { a = 1;//如果有括回来的括号，接收数据 } else { a = 0; } } } if (a==1) { total += num; } } printf("%d\n", total); } return 0; }

#include<stdio.h>
#include<string.h>
int main ()
{
    int n;
    scanf("%d",&n);
   for(int i=1;i<=n;i++)
   {
	int count=0;
	char ch[1000];
	memset(ch,'\0',sizeof(ch));//清空数组
	scanf("%s",ch);
	for(int m=0;;m++)
	{
		if(ch[m]=='(')//检测左括号
		{
			int t=m+1;
			while(ch[t]!=')')
			{t++;//记录括号内字符数量
			if(ch[t]=='\0')
			break;}
			if(ch[t]==')')//检测右括号
				{
					for(;m<t;m++)
					{
						if(ch[m]>='a'&&ch[m]<='z')//检测小写字母
		  				count++;
					}
				}
				if(ch[t]=='\0')
				{m=t;//标记跳出点
				break;}
		}
		if(ch[m]=='\0')//跳出循环
		break;
	}
	printf("%d\n",count);
   }
    return 0;
}

//仲夏特供
#include <iostream>
using namespace std;
int main() {
  string a;
  int T = 0;
  cin >> T;
  for (int i = 0; i < T; i++) {
    cin >> a;
    int length1 = a.length();
    int biaoJi1 = 0;
    int biaoJi2 = 0;
    for (int j = 0; j < length1; j++) {
      if (a[j] == '(') {
        biaoJi1 = j;
        break;
      }
    }
    for (int j = biaoJi1; j < length1; j++) {
      if (a[j] == ')') {
        biaoJi2 = j;
      }
    }
    int sum = 0;
    for (int j = biaoJi1 + 1; j < biaoJi2; j++) {
      //   cout << a[j] << endl;
      if ('a' <= a[j] && a[j] <= 'z') {
        sum++;
      }
    }
    cout << sum << endl;
  }
}