CF：Problem 383D - Antimatter 分治DP

参考：http://blog.youkuaiyun.com/u011466175/article/details/19689665

题目：http://codeforces.com/contest/383/problem/D

D. Antimatter
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 ai units of matter or ai 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, …, an (1 ≤ ai ≤ 1000).

The sum a1 + a2 + … + an 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

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 the i-th element produces antimatter.

题意：物质为正，反物质为负，组合起来就是+或者 - ，然后题目求的是：找出有多少种组合情况为0的数目（加起来为0）。

因为如果模拟组合情况的话，那么就是指数级别的了，肯定超时，所以只有DP了。
一个比较好的条件就是 sum<10000

using namespace std;  
typedef long long ll;  
int n, a;  
ll ans, d[2][22001]; //两种状态，即前一种状态的情况和当前状态情况，因为当前组合情况状态是在前一种组合情况状态下组合出来的  
int main()  
{  
    cin >> n;  
    for (int i = 0; n--; i ^= 1)  
    {  
        cin >> a;  
        //枚举每一种状态组合情况,几句就把+与-还有组合的情况就给模拟出来了，分治DP太强大了  
        for (int j = 1000; j <= 21000; ++j) //分治法，所以中间为11000  
            d[i][j]=(d[!i][j-a]+d[!i][j+a])%INF; //与其他数组合+或-的情况，j-a与j+a这思想即是分治，
        d[i][11000 + a]++; //自己为正的情况  
        d[i][11000 - a]++; //自己为负的情况  
        ans = (ans + d[i][11000]) % INF;  
    }  
    cout << ans << endl;  
    return 0;  
}