BZOJ 4870 组合数问题

Posted 2020-10-13 ziliuziliu

嗯。。。。这个式子很奇妙

化简是没有用的，考虑一个dp，答案能用这个式子表达。

于是dp[i][j]表示前i组物品选出%k=j个物品的方案数，然后矩阵加速转移。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
long long n,p,k,r,c[55][55];
struct matrix
{
    long long a[55][55];
}ans,base;
matrix operator * (matrix a,matrix b)
{
    matrix c;
    for (long long i=0;i<k;i++)
        for (long long j=0;j<k;j++)
            c.a[i][j]=0;
    for (long long i=0;i<k;i++)
        for (long long j=0;j<k;j++)
            for (long long kk=0;kk<k;kk++)
                c.a[i][j]=(c.a[i][j]+a.a[i][kk]*b.a[kk][j]%p)%p;
    return c;
}
void build()
{
    c[0][0]=1;
    for (long long i=1;i<=k;i++)
    {
        c[i][0]=1;
        for (long long j=1;j<=i;j++) c[i][j]=(c[i-1][j-1]+c[i-1][j])%p;
    }
    base.a[0][0]=2;for (long long i=1;i<k;i++) base.a[i][0]=c[k][k-i];
    for (long long i=1;i<k;i++)
        for (long long j=0;j<k;j++)
            base.a[j][i]=base.a[(j-1+k)%k][i-1];
    ans.a[0][0]=2;for (long long i=1;i<k;i++) ans.a[0][i]=c[k][i];
}
void f_pow(long long x)
{
    while (x)
    {
        if (x&1) ans=(ans*base);
        base=base*base;
        x>>=1;
    }
}
int main()
{
    scanf("%lld%lld%lld%lld",&n,&p,&k,&r);
    build();
    f_pow(n-1);
    printf("%lld\n",ans.a[0][r]);
    return 0;
}