【CUGBACM15级BC第23场 A】hdu 5146 Sequence

Sequence

[align=center]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1031    Accepted Submission(s): 583
[/align]

[align=left]Problem Description[/align]

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.
 

[align=left]Input[/align]

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
 

[align=left]Output[/align]

For each case output one line,if the sequence is good ,output "Yes",otherwise output "No".
 

[align=left]Sample Input[/align]

3
7
1 2 3 4 5 6 7
7
1 2 3 5 4 7 6
6
1 2 3 3 2 1

 

[align=left]Sample Output[/align]

No
Yes
No

 

题意：判断这个序列是不是好的，好的要求是在奇数坐标上的数的和等于在偶数坐标上的数的和且这个序列不是回文序列

#include<bits/stdc++.h>
#include <ctime>
using namespace std;
typedef long long ll;
const int MAXN = 1 * 1e5 + 500;
const ll M = 1e9 + 7;

int fun(int low, int high, int *str, int length)
{
if (length == 0 || length == 1)
{
return    1;
}
if (str[low] != str[high])
{
return    0;
}
return fun(low + 1, high - 1, str, length - 2);
}
int a[1500];
int main()
{
///clock_t start_time = clock();

///clock_t end_time = clock();
///cout << "Running time is: " << static_cast<double>(end_time - start_time) / CLOCKS_PER_SEC * 1000 << "ms" << endl;
std::ios::sync_with_stdio(false);
int t;
cin >> t;
while (t--)
{
int n, sum1 = 0, sum2 = 0;
cin >> n;
for (int i = 0; i < n; i++)
{
cin >> a[i];
if (i % 2 == 1)
{
sum1 += a[i];
}
else
{
sum2 += a[i];
}
}
//cout << sum1 << endl;
//cout << sum2 << endl;
if (fun(0, n - 1, a, n) == 1 || sum1 != sum2)
{
cout << "No" << endl;
}
else
{
cout << "Yes" << endl;
}
}
return 0;
}