hdu 5146(水题)

Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 924    Accepted Submission(s): 499

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

 

Source

BestCoder Round #23

 

题意:给一个串,如果这个串奇数位之和等于偶数位之和并且不是回文串输出Yes,否则输出No

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N = 1005;
int n,a[N];
bool test1(){
    LL sum1 = 0,sum2 = 0;
    for(int i=1;i<=n;i+=2){
        sum1+=a[i];
    }
    for(int i=2;i<=n;i+=2){
        sum2+=a[i];
    }
    if(sum1==sum2) return true;
    return false;
}
bool test2(){
    for(int i=1,j=n;i<j;i++,j--){
        if(a[i]!=a[j]) return true;
    }
    return false;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        if(test1()&&test2()) printf("Yes
");
        else printf("No
");
    }
    return 0;
}