Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 712    Accepted Submission(s): 439

Problem Description Today we have a number sequence A includes n elements.

Nero thinks a number sequence A is good only if the sum of its elements with odd index equals to the sum of its elements with even index and this sequence is not a palindrome.

Palindrome means no matter we read the sequence from head to tail or from tail to head,we get the same sequence.

Two sequence A and B are consider different if the length of A is different from the length of B or there exists an index i that
Ai≠Bi.
Now,give you the sequence A,check out it’s good or not.   Input The first line contains a single integer T,indicating the number of test cases.

Each test case begins with a line contains an integer n,the length of sequence A.

The next line follows n integers
A1,A2,…,An.

[Technical Specification]
1 <= T <= 100
1 <= n <= 1000
0 <= Ai <= 1000000   Output For each case output one line,if the sequence is good ,output “Yes”,otherwise output “No”.   Sample Input 3 7 1 2 3 4 5 6 7 7 1 2 3 5 4 7 6 6 1 2 3 3 2 1   Sample Output No Yes No  

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1005
const int inf=0x7fffffff;   //无限大
int a[maxn];
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        //memset(a,0,sizeof(a));
        int b;
        scanf("%d",&b);
        ll ans1=0;
        ll ans2=0;
        for(int i=0;i<b;i++)
        {
            scanf("%d",&a[i]);
            if(i%2==0)
                ans1+=a[i];
            else
                ans2+=a[i];
        }
        if(ans1!=ans2)
            cout<<"No"<<endl;
        else
        {
            int flag=0;
            for(int i=0;i<b;i++)
            {
                if(a[i]!=a[b-i-1])
                {
                    flag=1;
                    break;
                }
            }
            if(flag==0)
                cout<<"No"<<endl;
            else
                cout<<"Yes"<<endl;
        }
    }
    return 0;
}