poj2193

 1 //Accepted    368K    532MS
 2 //线性dp
 3 //dp[i][j]表示前i位最后一个是j的排列数
 4 //dp[i][j]=sum(dp[i-1][h])  h*2<=j
 5 #include <cstdio>
 6 #include <cstring>
 7 #include <iostream>
 8 using namespace std;
 9 const int imax_n = 15;
10 const int imax_m = 2005;
11 __int64 dp[imax_n][imax_m];
12 int n,m;
13 void Dp()
14 {
15     memset(dp,0,sizeof(dp));
16     for (int i=1;i<=m;i++)
17     dp[1][i]=1;
18     for (int i=2;i<=n;i++)
19     {
20         for (int j=1<<(i-1);j<=m;j++)
21         {
22             dp[i][j]=0;
23             for (int k=1<<(i-2);2*k<=j;k++)
24             {
25                 dp[i][j]+=dp[i-1][k];
26             }
27         }
28     }
29 }
30 int main()
31 {
32     int T;
33     scanf("%d",&T);
34     for (int t=1;t<=T;t++)
35     {
36         scanf("%d%d",&n,&m);
37         Dp();
38         __int64 ans=0;
39         for (int i=1<<(n-1);i<=m;i++)
40         ans+=dp[n][i];
41         printf("Case %d: n = %d, m = %d, # lists = %I64d\n",t,n,m,ans);
42     }
43     return 0;
44 }