hdu 4906 Our happy ending 状压dp

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http://acm.hdu.edu.cn/showproblem.php?pid=4906

Our happy ending

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1275 Accepted Submission(s): 447

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

12 2 2

Sample Output

Author

WJMZBMR

Source

2014 Multi-University Training Contest 4

题意：一个数量为n的序列，每个数0<=a[i]<=l，要求序列中的某些数之和等于k。求有多少种赋值方案。

题解：样例：可以赋值0 2，1 2，2 2，1 1，2 1，2 0。dp[i]：i转化为二进制，第一位表示和为1，第二位表示和为2……，二进制1表示有，0表示无。如i=3，二进制是11，表示有和为1，2的存在。下一个状态：next=(1<<(p-1))|i|(i<<k)&size。1、表示和为p的存在，2、i本身，3、有和为(i+p)存在。

代码：

#include<bits/stdc++.h>#define debug cout<<"aaa"<<endl#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson l,mid,root<<1#define rson mid+1,r,root<<1|1#define MIN_INT (-2147483647-1)#define MAX_INT 2147483647#define MAX_LL 9223372036854775807i64#define MIN_LL (-9223372036854775807i64-1)using namespace std;const int N = 100000 + 5;const int mod = 1000000000 + 7;int t,n,k,l,minn,v;LL dp[1<<21],ans;int main(){scanf("%d",&t);while(t--){int size,next;scanf("%d%d%d",&n,&k,&l);size=(1<<k)-1,minn=min(k,l),mem(dp,0),dp[0]=1,ans=0;for(int i=1;i<=n;i++){for(int j=size;j>=0;j--){if(dp[j]){v=dp[j];for(int p=1;p<=minn;p++){next=(1<<(p-1))|j|((j<<p)&size);dp[next]=(dp[next]+v)%mod;}if(k<l)dp[j]=(dp[j]+(LL)v*(l-k))%mod;}}}for(int i=0;i<=size;i++){if(i&(1<<(k-1))){ans=(ans+dp[i])%mod;}}printf("%lld\n",ans);}return 0;}