Codeforces 383D Antimatter【dp】

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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 devicei (1 ≤ i ≤ n) can create eithera_i units of matter ora_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 integersa₁, a₂, ..., a_n(1 ≤ a_i ≤ 1000).

The sum a₁ + a₂ + ... + a_n will be less than or equal to10000.

Output

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

Examples

Input

41 1 1 1

Output

Note

The possible photos are [1+, 2-], [1-, 2+], [2+, 3-], [2-, 3+], [3+, 4-], [3-, 4+], [1+, 2+, 3-, 4-], [1+, 2-, 3+, 4-], [1+, 2-, 3-, 4+], [1-, 2+, 3+, 4-], [1-, 2+, 3-, 4+] and [1-, 2-, 3+, 4+], where "i+" means that the i-th element produces matter, and "i-" means that thei-th element produces antimatter.

题目大意：

给你一个长度为N的序列，可以将任意数加上一个正号或者是负号.

求一共有多少个连续的子序列和为0.

思路：

1、统计计数问题，考虑dp：

①设定dp【i】【j】表示进行到第i位，和为j的方案数。

②考虑到会有负数和的情况出现，并且观察到题目保证a1+a2+a3+a4+................不会超过10000.那么显然，如果我们对所有数字都加上了负号，那么对应和不会小于-10000.

那么我们知道数组下标都是正的才行，那么我们考虑将15000设定为0.对应dp【i】【15000】就是表示加到第i位和为0的情况数。

2、那么接下我们考虑状态转移方程：

①dp【i】【j】+=dp【i-1】【j-a【i】】；

②dp【i】【j】+=dp【i-1】【j+a【i】】；

表示我们当前第i位的和紧接着上一位的和延续下去。

③dp【i】【15000+a【i】】=1；

④dp【i】【15000-a【i】】=1；

表示当前第i位的和独立出来，作为起点。

3、注意取模。

Ac代码：

#include<stdio.h>#include<string.h>using namespace std;#define mod 1000000007int a[1005];int dp[1005][30000];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        dp[1][15000+a[1]]=1;        dp[1][15000-a[1]]=1;        for(int i=2;i<=n;i++)        {            dp[i][15000+a[i]]=1;            dp[i][15000-a[i]]=1;            for(int j=0;j<30000;j++)            {                dp[i][j]=(dp[i][j]+dp[i-1][j-a[i]])%mod;                dp[i][j]=(dp[i][j]+dp[i-1][j+a[i]])%mod;            }        }        int output=0;        for(int i=1;i<=n;i++)        {            output=(output+dp[i][15000])%mod;        }        printf("%d\n",output);    }}

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