题目链接

  去看STDCALL的题解吧

  

#include<cstdio>
#include<cctype>
#define mod 9999973


inline long long read(){
    long long num=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')    f=-1;
        ch=getchar();
    }
    while(isdigit(ch)){
        num=num*10+ch-'0';
        ch=getchar();
    }
    return num*f;
}

int C(int num){
    return num*(num-1)/2;
}

long long f[102][102][102];
long long ans;
int main(){
    int n=read(),m=read();
    f[0][0][0]=1;
    for(int i=0;i<=n;++i)
        for(int j=0;j<=m;++j)
            for(int k=0;j+k<=m;++k){
                if(!f[i][j][k])    continue;
                f[i+1][j][k]=(f[i+1][j][k]+f[i][j][k])%mod;
                if(m-j-k>0)        f[i+1][j+1][k]=(f[i+1][j+1][k]+f[i][j][k]*(m-j-k))%mod;
                if(j)            f[i+1][j-1][k+1]=(f[i+1][j-1][k+1]+f[i][j][k]*j)%mod;
                if(m-j-k>1)        f[i+1][j+2][k]=(f[i+1][j+2][k]+f[i][j][k]*C(m-j-k))    %mod;
                if(m-j-k&&j)    f[i+1][j][k+1]=(f[i+1][j][k+1]+f[i][j][k]*(m-j-k)*j)%mod;
                if(j>1)            f[i+1][j-2][k+2]=(f[i+1][j-2][k+2]+f[i][j][k]*C(j))%mod;
            }
    for(int i=0;i<=m;++i)
        for(int j=0;i+j<=m;++j)
            ans=(ans+f[n][i][j])%mod;
    printf("%lld",ans);
    return 0;
}

 

  

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