bestcoder#23 1001 Sequence

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;

}