HDU6391 Lord Li's problem

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DP去重
AC Code:

#include<bits/stdc++.h>
using namespace std;
#define mod 19260817
inline void read(int &res){ char ch;for(;!isdigit(ch=getchar()););for(res=ch-'0';isdigit(ch=getchar());res=res*10+ch-'0'); }
int n,m;
char s[100],t[100];
int dp[41][21][61],fac[41],inv[41],invf[41];

inline int Pow(int base,int k)
{
    int ret=1;
    for(;k;k>>=1,base=1ll*base*base%mod) if(k&1) ret=1ll * ret * base % mod;
    return ret;
}

inline int C2(int n){ return n * (n-1) / 2; }
inline int C3(int n){ return n * (n-1) * (n-2) / 6; }

int main()
{
    fac[0] = fac[1] = inv[0] = inv[1] = invf[0] = invf[1] = 1;
    for(n=1;n<=40;n++)
    {
        if(n>1)
        {
            fac[n] = 1ll * fac[n-1] * n % mod;
            inv[n] = 1ll * (mod - mod / n) * inv[mod % n] % mod;
            invf[n] = 1ll * invf[n-1] * inv[n] % mod;
        }
        dp[n][0][0] = 1;
        for(int k=1;k<=min(20,n*(n-1)*(n-2)/6);k++)
            for(int j=0;j<=min(n,k*3);j++)
            {
                if(j+3 <= n) dp[n][k][j] = (dp[n][k][j] + 1ll * dp[n][k-1][j+3] * C3(j+3) % mod) % mod;
                if(j+1 <= n) dp[n][k][j] = (dp[n][k][j] + 1ll * dp[n][k-1][j+1] * C2(j+1) * (n-j-1) % mod)% mod;
                if(j-1 >= 0) dp[n][k][j] = (dp[n][k][j] + 1ll * dp[n][k-1][j-1] * C2(n-j+1) * (j-1) % mod)% mod;
                if(j-3 >= 0) dp[n][k][j] = (dp[n][k][j] + 1ll * dp[n][k-1][j-3] * C3(n - j + 3) % mod) % mod;
                if(k>=2) dp[n][k][j] = (dp[n][k][j] - 1ll * dp[n][k-2][j] * (k-1) % mod * (C3(n)-k+2) % mod + mod) % mod;
            }
    }
    int Case = 0;
    while(1)
    {
        ++Case;
        int cnt=0;
        read(n),read(m);
        if((!n) && (!m)) return 0;
        scanf("%s\n%s",s,t);
        for(int i=0;i<n;i++) s[i] = (s[i]-'0') ^ (t[i]-'0') , cnt+=(s[i]);
        printf("Case #%d: %lld\n",Case,1ll * dp[n][m][cnt] * fac[cnt] % mod * fac[n-cnt] % mod * invf[n] % mod * invf[m] % mod);
    }
}