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]

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<queue>
 7 #include<map>
 8 using namespace std;
 9 #define MOD 1000000007
10 const int INF=0x3f3f3f3f;
11 const double eps=1e-5;
12 typedef long long ll;
13 #define cl(a) memset(a,0,sizeof(a))
14 #define ts printf("*****\n");
15 const int MAXN=2005;
16 int n,m,tt;
17 int a[MAXN];
18 int dp[MAXN][MAXN];
19 int main()
20 {
21     int i,j,k,ca=1;
22     #ifndef ONLINE_JUDGE
23     freopen("1.in","r",stdin);
24     #endif
25     scanf("%d",&tt);
26     while(tt--)
27     {
28         scanf("%d",&n);
29         int sum=0;
30         for(i=0;i<n;i++)
31         {
32             scanf("%d",a+i);
33             sum+=a[i];
34         }
35         cl(dp);
36         for(i=0;i<=a[0];i++)
37         {
38             dp[0][i]=1;
39         }
40         for(i=1;i<n;i++)
41         {
42             for(j=0;j<=sum/2;j++)
43             {
44                 for(k=0;k<=a[i];k++)
45                 {
46                     if(j-k<0)   break;
47                     dp[i][j]+=dp[i-1][j-k];
48                     dp[i][j]%=MOD;
49                 }
50             }
51         }
52         printf("%d\n",dp[n-1][sum/2]);
53     }
54 }