hdu 5000 dp **

题目中提到  It guarantees that the sum of T[i] in each test case is no more than 2000 and 1 <= T[i].

加上一堆顺序，基本可以猜到是属于递推形的dp

dp[i][j]表示前i个和为j时的方案数

dp方程:   dp[i][j]+=dp[i][j-k]

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
using namespace std;
#define MOD 1000000007
const int INF=0x3f3f3f3f;
const double eps=1e-5;
typedef long long ll;
#define cl(a) memset(a,0,sizeof(a))
#define ts printf("*****
");
const int MAXN=2005;
int n,m,tt;
int a[MAXN];
int dp[MAXN][MAXN];
int main()
{
    int i,j,k,ca=1;
    #ifndef ONLINE_JUDGE
    freopen("1.in","r",stdin);
    #endif
    scanf("%d",&tt);
    while(tt--)
    {
        scanf("%d",&n);
        int sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",a+i);
            sum+=a[i];
        }
        cl(dp);
        for(i=0;i<=a[0];i++)
        {
            dp[0][i]=1;
        }
        for(i=1;i<n;i++)
        {
            for(j=0;j<=sum/2;j++)
            {
                for(k=0;k<=a[i];k++)
                {
                    if(j-k<0)   break;
                    dp[i][j]+=dp[i-1][j-k];
                    dp[i][j]%=MOD;
                }
            }
        }
        printf("%d
",dp[n-1][sum/2]);
    }
}