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题目链接:
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + ‘0‘);
putchar(‘\n‘);
}
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e5+10;
const int maxn=5e3+4;
const double eps=1e-12;
int n;
char s[N];
int check()
{
int num1=0,num2=0;
For(i,1,n)
{
if(s[i]==‘(‘)num1++;
else num2++;
}
if(num1!=num2)return 0;
if(n==2)
{
if(s[1]==‘(‘&&s[2]==‘)‘)return 0;
return 1;
}
num1=0;num2=0;
For(i,1,n)
{
if(s[i]==‘(‘)num1++;
else
{
if(num1==0)num2++;
else num1--;
}
}
if(num2>2)return 0;
return 1;
}
int main()
{
int T;
read(T);
while(T--)
{
read(n);
scanf("%s",s+1);
if(check())printf("Yes\n");
else printf("No\n");
}
return 0;
}
hdu-5831 Rikka with Parenthesis II(贪心)
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5762208.html