题目示例：

分析

由与没有给出字符串的长度，所以只能通过getline一次性处理，而在输入后恰好能倒序处理字符串，以标点符号为分界点，将数字当成字符放到栈里，遇到下一个标点符号时执行查找操作，并清空栈。

题不难，直接上代码实现吧：

//程序没考虑空格。请用这行没空格的样例 
//样例： matrix=[[1,3,5,7],[10,11,16,20],[23,30,34,60]],target=3
//matrix=[[1,3,5,7],[10,11,16,20],[23,30,34,60]],target=11#include <iostream>
#include <string>
#include <cstring>
#include <stack>
using namespace std;stack<char> c;char tar[1000];bool ans;void del()
{while (!c.empty()){c.pop();}
}bool equ(int t, char ss[], char s[])
{for (int i = 0; i<=t ; i++){if (s[i] != ss[i]){return false;}}return true;
}bool com()
{char b[10000]={'\0'};int i = 0;while (!c.empty()){b[i] = c.top();c.pop();i++;
//		cout<<"cbb "<<b[i]<<" this"<<endl;}//	cout<<"ans"<<tar<<" "<<b<<endl;if (equ(i, b, tar)){ans = true;//		cout<<"ans"<<tar<<" "<<b<<endl;return ans;}return false;
}int main()
{string a;getline(cin,a);int n = a.size();int x = n;char temp = a[n-1];while (temp != '='){n--;temp = a[n-1];
//		cout<<n<<endl;} for (int i = n, j = 0;i <= x-1; i++,j++){tar[j] = a[i];
//		cout<<"t"<<a[i]<<endl;}//提取目标字符串 //	cout<<"tar"<<tar<<endl;n--;//减去等号n--;//再减1直接拿n当下标n-=6;//减去target //	cout<<"a[n]"<<a[n]<<endl;while (a[n] != '='){if (a[n] == ']'){n--;continue;} if (a[n] == ',' || a[n] == '['){
//			cout<<"where"<<endl;if (!c.empty()){if (com()){break;}}n--;continue;}c.push(a[n]);//		cout <<"n"<< n<<" "<<a[n]<<" ";n--;} if (ans == true){cout<<"true";}else{cout<<"false";}return 0;
}