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HDU 4906

2024-07-15 11:28:31   214人阅读

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
12 2 2
 

 

Sample Output
6


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，表示前i个数的和（1-20）是否达到 

 

#include "cstdio"#include "cstring"#define min(x,y) (x>y?y:x)#define MOD 1000000007int f[1050000];int n,k,l;int main(){    int tt;    scanf("%d",&tt);    while(tt--)    {        memset(f,0,sizeof(f));        scanf("%d%d%d",&n,&k,&l);        int ms=(1<<k)-1;        long long ans=0;        for(int j=0; j<=min(l,k); j++)            f[1<<(j-1)]=1;        if(l>k) f[0]=(f[0]+l-k)%MOD;        for(int i=1; i<n; i++)            for(int s=ms; s>=0; s--)            if(f[s]>0)//            {                long long tmp=f[s];                for(int j=1; j<=min(l,k); j++)                {                    int ns=(s|(s<<j)|(1<<(j-1)))&ms;                    f[ns]=(f[ns]+tmp)%MOD;                }                if(l>k)                    f[s]=(f[s]+(l-k)*tmp%MOD)%MOD;            }        for(int s=0; s<=ms; s++)        if(s&(1<<(k-1)))            ans=(ans+f[s])%MOD;        printf("%lld\n",ans);    }    return 0;}
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超时代码：

#include "cstdio"#include "cstring"#define min(x,y) (x>y?y:x)#define MOD 1000000007int f[21][1050000],c[21][21];int n,k,l;inline long long pow(long long a, long long b){    long long ret=1,tmp=a%MOD;    while(b)    {        if(b&1) ret=(ret*tmp)%MOD;        tmp=(tmp*tmp)%MOD;        b>>=1;    }    return ret;}int main(){    for(int i=0; i<=20; i++)        for(int j=0; j<=i; j++)        if(j>0) c[i][j]=c[i-1][j]+c[i-1][j-1];        else c[i][j]=1;    int tt;    scanf("%d",&tt);    while(tt--)    {        memset(f,0,sizeof(f));        scanf("%d%d%d",&n,&k,&l);        int ms=(1<<k)-1;        for(int j=0; j<=min(l,k); j++)            f[1][1<<(j-1)]=1;        for(int i=1; i<n; i++)            for(int j=0; j<=min(l,k); j++)                for(int s=0; s<=ms; s++)                {                    f[i+1][(s|(s<<j)|(1<<(j-1)))&ms]+=f[i][s];                    f[i+1][(s|(s<<j)|(1<<(j-1)))&ms]%=MOD;                }        long long ans=0;        if(l<=k)        {            for(int s=0; s<=ms; s++)            if(s&(1<<(k-1)))                ans=(ans+f[n][s])%MOD;        }        else        {            for(int i=1; i<=n; i++)                for(int s=0; s<=ms; s++)                if(s&(1<<(k-1)))                    ans=(ans+((long long)f[i][s]*(long long)c[n][i])%MOD*pow(l-k,n-i))%MOD;        }        printf("%lld\n",ans);    }    return 0;}
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