CF918C The Monster

题目链接：http://codeforces.com/contest/918/problem/C

知识点：　　贪心

解题思路：

　　枚举起点（当起点就是\(')'\)时直接跳过）并在此基础上遍历字符串，用一个\(nowmin\)和一个\(nowmax\)来记录当前\('('\)最多有\(nowmax\)个，最少有\(nowmin\)个。当遍历到\('('\)时,\(nowmin++,nowmax++\)；当遍历到\('?'\)时，\(nowmin--,nowmax++\)（因为\('?'\)既有可能是\('('\)，也有可能是\(')'\)）；当遍历到\(')'\)时，\(nowmin++,nowmax++\)。当\(nowmax<0\)时就可以结束对这个起点的字符串的遍历，当\(nowmin<0\)时将其置零，当\(nowmin==0\)并且\(nowmax>=0\)并且当前遍历长度为偶数时，答案数加一。

　　思路来自magicalCycloidea.

AC代码：

 #include <bits/stdc++.h>

 using namespace std;
 const int maxn = ;
 char inp[maxn];
 int main()
 {
     scanf("%s",inp);
     int len=strlen(inp);
     int ans=;
     for(int i=;i<len;i++){
         if(inp[i]==')') continue;
         int nowmin=,nowmax=;
         for(int j=i,l=;j<len;j++,l++){
             if(inp[j]=='(') nowmax++,nowmin++;
             else if(inp[j]=='?')    nowmax++,nowmin--;
             else    nowmax--,nowmin--;
             if(nowmax<)    break;
             nowmin=max(,nowmin);
             if(l%==&&nowmin==&&nowmax>=)    ans++;
         }
     }
     printf("%d\n",ans);
     return ;
 }