hdu 4906 Our happy ending 。神奇的状态转移方程，记录下

转自http://www.cnblogs.com/20120125llcai/p/3883896.html

Our happy ending

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 194 Accepted Submission(s): 48



Problem Description

There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil.
Also, this devil is looking like a very cute Loli.

Y*wan still remember the day he first meets the devil. Now everything is done and the devil is gone. Y*wan feel very sad and suicide.

You feel guilty after killing so many loli, so you suicide too.

Nobody survive in this silly story, but there is still some hope, because this is just a silly background story during one programming contest!

And the last problem is:

Given a sequence a_1,a_2,...,a_n, if we can take some of them(each a_i can only be used once), and they sum to k, then we say this sequence is a good sequence.

How many good sequence are there? Given that each a_i is an integer and 0<= a_i <= L.

You should output the result modulo 10^9+7.

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains 3 integers n, k, L.

T<=20, n,k<=20 , 0<=L<=10^9.

Output

For each cases, output the answer in a single line.

Sample Input

1
2 2 2

Sample Output

6

Author

WJMZBMR

算是状态dp吧

dp[i] 表示 i 状态下的方案数。

i 化为二进制时 ，第一位为 1 表示有和为1 的存在，0表示没有，第二位为 1 时表示有和为 2的存在

第三位为 1 时表示有和为 3的存在...

转移：sta = i|(1<<(k-1))|(i<<j&m) ；

表示 i 的状态转移到了 sta状态

假如当前的状态 为 3 二进制就是 11,表示有和为 1,2的存在，

如果加上 3 ，那么 3 的状态就可以达到，也就是 |(1<<(3-1))

还有3可以和1,2加，也就是 原来的加上 3 ，也就是左移 3 位 |(i<<j)&m

因为大于 k的和我们是不在意的，所以最大的状态就是 (1<<k)-1 ;

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define maxn 100010
#define mod 1000000007
#define LL long long
using namespace std;
int dp[(1<<20)+10] ;
int main()
{
int i ,n,m,k,j ;
int T ,L ;
cin >> T ;
while( T-- )
{
scanf("%d%d%d",&n,&k,&L) ;
memset(dp,0,sizeof(dp)) ;
dp[0]=1 ;
m = (1<<k)-1 ;
int tot = 0 ;
if(L > k)
{
tot = L-k ;
L = k ;
}
while(n--)
{
for( i = m ; i >= 0 ;i--)
{
if(dp[i]==0) continue ;
j = dp[i] ;
LL tmp = (LL)tot*dp[i]%mod ;
for( int u = 1 ; u <= L ;u++)<span style="white-space:pre">		</span>//加上一个小于等于k的数
{
int sta = (i|(1<<(u-1))|((i<<u)&m)) ;
dp[sta] += j ;
dp[sta] %= mod;
}
dp[i] = (dp[i]+tmp)%mod ;<span style="white-space:pre">		</span>//加上一个大于k的数
}
}
LL ans = 0 ;
for( i = 0 ; i <= m ;i++)
if((i>>(k-1))&1)
{
ans = (ans+dp[i])%mod ;
}
cout << ans << endl;
}
return 0 ;
}