CodeForces 383D Antimatter

Iahub accidentally discovered a secret lab. He found there n devices ordered in a line, numbered from 1 to n from left to right. Each device i (1 ≤ i ≤ n) can create either a_i units of matter or a_i units of antimatter.

Iahub wants to choose some contiguous subarray of devices in the lab, specify the production mode for each of them (produce matter or antimatter) and finally take a photo of it. However he will be successful only if the amounts of matter and antimatter produced in the selected subarray will be the same (otherwise there would be overflowing matter or antimatter in the photo).

You are requested to compute the number of different ways Iahub can successful take a photo. A photo is different than another if it represents another subarray, or if at least one device of the subarray is set to produce matter in one of the photos and antimatter in the other one.

Input

The first line contains an integer n (1 ≤ n ≤ 1000). The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 1000).

The sum a₁ + a₂ + ... + a_n will be less than or equal to 10000.

Output

Output a single integer, the number of ways Iahub can take a photo, modulo 1000000007 (10⁹ + 7).

Sample test(s)

Input

4
1 1 1 1

Output

12

提意：给出一串序列，求出序列满足连续且相加和为0的个数。
sl:背包问题，但是需要在每次以i为末尾节点后更新这个点的初始值把它作为开始节点考虑进去。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int MAX = 1e3+10;
 6 const int low = 3*10000;
 7 const int MOD = 1e9+7;
 8 const int M = 10000+10;
 9 int a[MAX];
10 int dp[MAX][2*low];
11 int main()
12 {
13     int n,ans,sum;
14     while(scanf("%d",&n)==1)
15     {
16         sum=0;
17         for(int i=0;i<n;i++) scanf("%d",&a[i]);
18         memset(dp,0,sizeof(dp));
19         dp[0][low]=1;
20         for(int i=1;i<=n;i++)
21         {
22             for(int j=-M;j<=M;j++)
23             {
24                 dp[i][j+low]=(dp[i-1][j+low+a[i-1]]+dp[i-1][j+low-a[i-1]])%MOD;
25             }
26             sum=(sum+dp[i][low])%MOD;
27             dp[i][low]++;
28         }
29         printf("%d
",sum%MOD);
30     }
31 }