hdu 3944 lucas

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需要分情况讨论。

然后求组合数套上lucas。

#include <bits/stdc++.h>using namespace std;typedef long long ll;ll n, k, p;ll powmod(ll a, ll b){ ll res = 1; a %= p; while(b){ if(b&1) res = res*a%p; b>>=1; a = a*a%p; } return res; }ll fac[1310][10100];bool is[100010];int cnt;int pr[10010];int id[10010];void init(){    cnt = 0;    for(int i=2; i<=10000; i++){        if(!is[i]){            pr[++cnt] = i;            id[i] = cnt;            for(int j=i+i; j<=10000; j+=i)                is[j] = true;        }    }    for(int i=1; i<=cnt; i++){        fac[i][0] = 1;        for(int j=1; j<=10000; j++){            fac[i][j] = fac[i][j-1] * j % pr[i];         }    }}ll getc(int n, int m, ll p, int tmp){    if(m > n) return 0;    ll res = fac[tmp][n]*powmod(fac[tmp][n-m]*fac[tmp][m]%p, p-2)%p;    return res;}ll lucas(int n, int m, ll p, int tmp){    if(m == 0) return 1;    return lucas(n/p, m/p, p, tmp)*getc(n%p, m%p, p, tmp)%p;}int main(){    int icase = 0;    init();    while(~scanf("%lld %lld %lld", &n, &k, &p)){        int tmp = id[p];        if(k > n/2) k = n- k;        ll res = lucas(n+1, k, p, tmp);        printf("Case #%d: %lld\n", ++icase, (res + n - k)%p);    }    return 0;}

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