HDU 6391 Lord Li's problem 组合数学

题目链接：https://vjudge.net/problem/HDU-6391

思路：转载自https://blog.csdn.net/qq_34454069/article/details/81669398

AC代码：

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

#define mod 19260817
using ll=long long;
ll dp[43][23][43];//n个数字，k次选择，m次为1
ll n,k;
ll fac[43];
ll ifac[43];
ll inv[43];

long long quick_pow(long long a,long long t)
{
    long long ans=1;
    while(t)
    {
        if(t&1)
            ans=ans*a%mod;
        a=a*a%mod;
        t=t/2;
    }
    return ans;
}

void init()
{
    fac[0]=fac[1]=1;
    for(int i=2;i<=40;++i)
        fac[i]=i*fac[i-1]%mod;
    ifac[40]=quick_pow(fac[40],mod-2);
    for(int i=39;i>=0;--i)
        ifac[i]=ifac[i+1]*(i+1)%mod;
    inv[0]=inv[1]=1;
    for(int i=2;i<=40;++i)
        inv[i]=(mod-mod/i)*inv[mod%i]%mod;
}

long long C(int x,int y)
{
    if(x<y)
        return 0;
    return fac[x]*ifac[y]%mod*ifac[x-y]%mod;
}


int main()
{
    init();
    for(int i=1;i<=40;++i)
    {
        dp[i][0][0]=1;
        for(int j=1;j<=min((long long)20,C(i,3));++j)
        {
            for(int k=0;k<=i;++k)
            {
                if(k-3>=0) dp[i][j][k]=(dp[i][j-1][k-3]*C(i-k+3,3)%mod+dp[i][j][k])%mod;
                if(k-1>=0) dp[i][j][k]=(dp[i][j-1][k-1]*C(i-k+1,2)%mod*C(k-1,1)%mod+dp[i][j][k])%mod;
                if(k+3<=i) dp[i][j][k]=(dp[i][j-1][k+3]*C(k+3,3)%mod+dp[i][j][k])%mod;
                if(k+1<=i) dp[i][j][k]=(dp[i][j-1][k+1]*C(k+1,2)%mod*C(i-k-1,1)%mod+dp[i][j][k])%mod;
                if(j-2>=0) dp[i][j][k]=(dp[i][j][k]+mod-(dp[i][j-2][k]*(j-1)%mod*(C(i,3)-j+2))%mod)%mod;
            }
        }
    }
    int ca=1;
    while(scanf("%d %d",&n,&k)&&(n||k))
    {
        char str1[43];
        char str2[43];
        scanf("%s",str1);
        scanf("%s",str2);
        int m=0;
        for(int i=0;i<n;++i)
            m+=(str1[i]-'0')^(str2[i]-'0');
        cout<<"Case #"<<ca++<<": ";
        cout<<dp[n][k][m]*quick_pow(C(n,m),mod-2)%mod*ifac[k]%mod<<endl;
    }

    return 0;
}