lucas求组合数C(n,k)%p

Saving Beans http://acm.hdu.edu.cn/showproblem.php?pid=3037

#include<cstdio>
typedef __int64 LL;
const int M=100010;
class LUCAS { //lucas求组合数C(n,k)%p
    LL F[M];
    LL inv(LL a,LL mod) {
        if(a==1) return 1;
        return inv(mod%a,mod)*(mod-mod/a)%mod;
    }
    void init(LL p) {
        F[0]=1;
        for(int i=1; i<=p; i++) {
            F[i]=F[i-1]*i%p;
        }
    }
    LL Lucas(LL n,LL k,LL p) {
        LL ans=1;
        while(n&&k) {
            LL a=n%p;
            LL b=k%p;
            if(a<b) return 0;
            ans=ans*F[a]%p*inv(F[b]*F[a-b]%p,p)%p;
            n/=p;
            k/=p;
        }
        return ans;
    }
public:
    LL solve(LL n,LL k,LL p){
        init(p);
        return Lucas(n,k,p);
    }
}gx;
int main() {
    int t,n,m,p;
    while(~scanf("%d",&t)) {
        while(t--) {
            scanf("%d%d%d",&n,&m,&p);
            printf("%d
",(int)gx.solve(n+m,n,p));
        }
    }
    return 0;
}

DP? http://acm.hdu.edu.cn/showproblem.php?pid=3944

#include<cstdio>
#include<cstring>
#include<vector>
#define mt(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef __int64 LL;
const int M=10010;
class LUCASM { //比较适合n,k<=10^9,p<=10^4的情况
    int flag[M*10],prime[1300],pcnt;
    vector<int> rev[M],fac[M];
    int quickpow(int a,int b,int c) { //快速幂求(a^b)%c
        int ret=1%c;
        for(; b; a=a*a%c,b>>=1) {
            if(b&1) {
                ret=ret*a%c;
            }
        }
        return ret;
    }
public:
    void init() {//先初始化一次即可
        pcnt=0;
        mt(flag,-1);
        for(int i=2; i<=10000; i++) {
            if(flag[i]) {
                prime[pcnt++]=i;
                rev[i].clear();
                fac[i].clear();
            }
            for(int j=0; j<pcnt&&prime[j]<=10000/i; j++) {
                flag[i*prime[j]]=0;
                if(!(i%prime[j])) break;
            }
        }
        for(int i=0; i<pcnt; i++) {
            int tnum=1;
            rev[prime[i]].push_back(1);
            fac[prime[i]].push_back(1);
            for (int j=1; j<prime[i]; j++) {
                tnum=(tnum*j)%prime[i];
                int now=quickpow(tnum,prime[i]-2,prime[i]);
                fac[prime[i]].push_back(tnum);
                rev[prime[i]].push_back(now);
            }
        }
    }
    int lucas(int n,int k,int p) {
        int ret=1;
        while (n && k) {
            int num1=n%p;
            int num2=k%p;
            n/=p;
            k/=p;
            if (num1<num2) return 0;
            int num=(fac[p][num1]*rev[p][num2])%p;//计算c(num1,num2)%p
            num=(num*rev[p][num1-num2])%p;
            ret=(ret*num)%p;
        }
        return ret;
    }
} gx;
int main() {
    int n,k,p,cas=1;
    gx.init();
    while(~scanf("%d%d%d",&n,&k,&p)) {
        printf("Case #%d: ",cas++);
        if(k>n/2) k=n-k;
        int o=gx.lucas(n+1,k,p);
        printf("%d
",(n-k+o)%p);
    }
    return 0;
}

end