| ||||||||||
| Online Judge | Problem Set | Authors | Online Contests | User | ||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Web Board Home Page F.A.Qs Statistical Charts | Current Contest Past Contests Scheduled Contests Award Contest | |||||||||
记录一下。#include <iostream>
using namespace std;
int GetLeftNum(int index,int s[],bool vis[]);
int main()
{
int N;
cin>>N;
while(N--)
{
int num;
cin>>num;
int *P = new int[num];
//int *W = new int[num];
for(int i = 0;i<num;++i)
cin>>P[i];
int *str = new int[num + P[num - 1]]; //即为括号串的长度
bool *vis = new bool[num + P[num - 1]]; //辅助访问数组
for(int i = 0;i<num + P[num - 1];++i) //初始化数组
{
str[i] = 0;
vis[i] = false;
}
for(int i = 0;i<num;++i)
{
str[i + P[i]] = 1; //该位置为右括号1
cout<<GetLeftNum(i+P[i]-1,str,vis)<<" ";
}
cout<<endl;
}
return 0;
}
int GetLeftNum(int index,int s[],bool vis[])
{
int count = 0;
for(;index>=0;index--)
{
if(s[index] == 0)
{
count++;
if(!vis[index])
{
vis[index] = true;
break;
}
}
}
return count;
}
Followed by: Post your reply here: |
All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator