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 = 5005;
char inp[maxn];
int main()
{
    scanf("%s",inp);
    int len=strlen(inp);
    int ans=0;
    for(int i=0;i<len;i++){
        if(inp[i]==')') continue;
        int nowmin=0,nowmax=0;
        for(int j=i,l=1;j<len;j++,l++){
            if(inp[j]=='(') nowmax++,nowmin++;
            else if(inp[j]=='?')    nowmax++,nowmin--;
            else    nowmax--,nowmin--;
            if(nowmax<0)    break;
            nowmin=max(0,nowmin);
            if(l%2==0&&nowmin==0&&nowmax>=0)    ans++;
        }
    }
    printf("%d\n",ans);
    return 0;
}