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-;
 typedef long long ll;
 #define cl(a) memset(a,0,sizeof(a))
 #define ts printf("*****\n");
 const int MAXN=;
 int n,m,tt;
 int a[MAXN];
 int dp[MAXN][MAXN];
 int main()
 {
     int i,j,k,ca=;
     #ifndef ONLINE_JUDGE
     freopen("1.in","r",stdin);
     #endif
     scanf("%d",&tt);
     while(tt--)
     {
         scanf("%d",&n);
         int sum=;
         for(i=;i<n;i++)
         {
             scanf("%d",a+i);
             sum+=a[i];
         }
         cl(dp);
         for(i=;i<=a[];i++)
         {
             dp[][i]=;
         }
         for(i=;i<n;i++)
         {
             for(j=;j<=sum/;j++)
             {
                 for(k=;k<=a[i];k++)
                 {
                     if(j-k<)   break;
                     dp[i][j]+=dp[i-][j-k];
                     dp[i][j]%=MOD;
                 }
             }
         }
         printf("%d\n",dp[n-][sum/]);
     }
 }