HDU 5146 Sequence 判断序列不是回文 且奇数与偶数位的和不同

Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 173    Accepted Submission(s): 126

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 thatAi≠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

371 2 3 4 5 6 771 2 3 5 4 7 661 2 3 3 2 1

Sample Output

NoYesNo

/*5146 判断序列不是回文 且奇数与偶数位的和不同 */#include<iostream>#include<stdio.h>using namespace std;int a[1001];int main(){int t,n,sum,i,j,f;scanf("%d",&t);while(t--){scanf("%d",&n);sum=0;for(i=0;i<n;i++){scanf("%d",&a[i]);if(i&1==1)sum+=a[i];elsesum-=a[i];}f=0;for(i=0,j=n-1;i<=j;i++,j--){if(a[i]!=a[j]){f=1;break;}}if(sum==0&&f==1)printf("Yes\n");elseprintf("No\n");}return 0;}