HDU 5146 Sequence 回文数组

原题： http://acm.hdu.edu.cn/showproblem.php?pid=5146

题目：

Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 762 Accepted Submission(s): 396

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

思路：

判断一串数字奇数位置上的元素是否为总和的一半，并且判断该数串是否回文。

代码：

#include <iostream>#include"cstdio"#include"string.h"#include"vector"using namespace std;const int N = 1005;int a[N];int n,m;int main(){    int tt;    scanf("%d",&tt);    while(tt--)    {        int sum1=0;        int sum=0;        scanf("%d",&n);        for(int i=1; i<=n; i++)        {            scanf("%d",&a[i]);            if(i%2==1)  sum1=sum1+a[i];            sum=sum+a[i];        }        int flag=1;        if(sum1*2==sum)        {            for(int i=1,j=n; i<=j; i++,j--)                if(a[i]!=a[j])                {                    flag=0;                    break;                }        }        if(flag==0)    printf("Yes\n");        else printf("No\n");    }}