HDU 4906 状态压缩dp

Our happy ending

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 570    Accepted Submission(s): 183

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

 

题意：求n个数组成的有序数列，满足存在一个子数列和为k，且每个数在[0,l]的范围内，这样的有序数列有多少个？

递推实现 f[i][j]表示前i个数能够表示j状态的方案数，其中j为最多20位的二进制，每一二进制位表示‘和’为该位置是否达到，比如第20位二进制位若为1，表示的就是前i个数和有达到20，这20位表示的就是一个状态。

从j状态向上推，若该位置取k，则 需要更新的状态为 j | j<<k | 1<<(k-1) 注意包括0位置

可以用一维数组实现滚动数组优化

复杂度理论上高达O(n^2*2^k)会超时，但是要注意到很大部分f[i][j]=0，加上这个优化瞬间快了10倍多

　　　　中间的递推过程从标程中照搬下来了，惭愧！

#include "cstdio"
#include "cstring"
#define min(x,y) (x>y?y:x)

#define MOD 1000000007

int f[];
int n,k,l;
int main()
{
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        memset(f,,sizeof(f));
        scanf("%d%d%d",&n,&k,&l);
        int ms=(<<k)-;
        long long ans=;

        for(int j=; j<=min(l,k); j++)
            f[<<(j-)]=;
        if(l>k) f[]=(f[]+l-k)%MOD;

        for(int i=; i<n; i++)
            for(int s=ms; s>=; s--)    //从大到小 避免重复计算
            if(f[s]>)    //没加这句20s+ 加了就1286ms
            {
                long long tmp=f[s];        //滚动数组压缩 发现滚动数组都不需要了 !
                for(int j=; j<=min(l,k); j++)    //f[i][s] 插入j=0的情况直接保存给 f[i+1][s]
                {
                    int ns=(s|(s<<j)|(<<(j-)))&ms;    //包括从0处向右推j位
                    f[ns]=(f[ns]+tmp)%MOD;
                }
                if(l>k)
                    f[s]=(f[s]+(l-k)*tmp%MOD)%MOD;
            }

        for(int s=; s<=ms; s++)
        if(s&(<<(k-)))
            ans=(ans+f[s])%MOD;

        printf("%lld\n",ans);
    }
    return ;
}

之前看解题报告写的

　　　　开了二维数组，画蛇添足的加了个快速幂和组合 =.=  时间20s+

超时代码：

#include "cstdio"
#include "cstring"
#define min(x,y) (x>y?y:x)
#define MOD 1000000007
int f[][],c[][];
int n,k,l;
inline long long pow(long long a, long long b)
{
    long long ret=,tmp=a%MOD;
    while(b)
    {
        if(b&) ret=(ret*tmp)%MOD;
        tmp=(tmp*tmp)%MOD;
        b>>=;
    }
    return ret;
}
int main()
{
    for(int i=; i<=; i++)
        for(int j=; j<=i; j++)
        if(j>) c[i][j]=c[i-][j]+c[i-][j-];
        else c[i][j]=;
    int tt;
    scanf("%d",&tt);
    while(tt--)
    {
        memset(f,,sizeof(f));
        scanf("%d%d%d",&n,&k,&l);
        int ms=(<<k)-;
        for(int j=; j<=min(l,k); j++)
            f[][<<(j-)]=;
        for(int i=; i<n; i++)
            for(int j=; j<=min(l,k); j++)
                for(int s=; s<=ms; s++)
                {
                    f[i+][(s|(s<<j)|(<<(j-)))&ms]+=f[i][s];
                    f[i+][(s|(s<<j)|(<<(j-)))&ms]%=MOD;
                }
        long long ans=;
        if(l<=k)
        {
            for(int s=; s<=ms; s++)
            if(s&(<<(k-)))
                ans=(ans+f[n][s])%MOD;
        }
        else
        {
            for(int i=; i<=n; i++)
                for(int s=; s<=ms; s++)
                if(s&(<<(k-)))
                    ans=(ans+((long long)f[i][s]*(long long)c[n][i])%MOD*pow(l-k,n-i))%MOD;
        }
        printf("%lld\n",ans);
    }
    return ;
}