杭电5146题

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

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

 

Sample Output

NoYesNo

 

（就是判断一个数列的奇数项和与偶数项和是否相等，若相等且不是回文数列，则为好的数列）

#include <iostream>using namespace std;int x[1000000];int main(){int m,n,s1,s2;int i,j;cin >> m;while (m--){cin >> n;s1 = s2 = 0;for (i = 0; i < n; i++) {cin >> x[i];if(( i + 1 ) % 2 == 0) s2 += x[i];else s1 += x[i];}//判断数列奇数项和偶数项之和是否相等if(s1 == s2){s1 = s2 =0;for(i = 0, j = n - 1; i <= n / 2; i++, j--){s1 += x[i];s2 += x[j];} //判断是否为回文数列if(s1 == s2) cout << "No" << endl;else cout << "Yes" << endl;}else cout << "No" << endl;}return 0;}

（鄙人还是新手，代码写的比较烂。欢迎各位高手留言，对代码精简完善）