1116 Necklace
时间限制：1000ms 内存限制：32000KB java 两倍。
介绍
Little King has a beautiful pearl necklace, one day, he find that there is a number in each pearl, so he want to know whether he can find a continuous sequence that the sum of them is the number he give to you.
输入格式描述
The first line contains one integer T(T<=50),indicating the number of test cases.
For each test case ,the first line contains two integer N(5<=N<=100000),K(1<=K<=109)(within int),indicating there are N pearls in necklace and the number he give to you .The second line contains N integer Ai(1<=Ai<=10000),indicating the number in each pearl, pearls are sort by clockwise.
输出格式描述
For each test case, if he can find out print YES, otherwise print NO.
5 15
1 2 3 4 5
5 16
1 2 3 4 5
6 18
1 2 3 4 5 7
NO
YES
突然想起来同学学校的题目我还没写。刚开始纠结于第三个例子，后来问了同学才想起来这是个项链，这坑会造成数组越界，稍微处理一下。怎么把cin同步关掉快scanf这么多，奇怪
代码：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
int list[300010];
int main (void)
{
ios::sync_with_stdio(false);
int t,i,j,n,s;
cin>>t;
while (t--)
{
memset(list,0,sizeof(list));
cin>>n>>s;
for (i=1; i<=n; i++)
{
cin>>list[i];
list[n+i]=list[i];
}
bool flag=false;
int l=1,r=1,temp=0;
while (1)
{
while (temp<s&&r<=2*n)
{
temp+=list[r++];
}
if(temp<s)
break;
if(temp==s&&r-l<=n)
{
flag=true;
break;
}
temp-=list[l++];
if(temp==s&&r-l<=n)
{
flag=true;
break;
}
}
if(flag)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
}
return 0;
}