组合c(m,n)的计算方法

int Combination(int n, int m) 
{ const int M = 10007; int ans = 1; for(int i=n; i>=(n-m+1); --i) ans *= i; while(m) ans /= m--; return ans % M; 
} int Combination(int n, int m)
{const int M = 10007;int ans = 1;for(int i=n; i>=(n-m+1); --i)ans *= i;while(m)ans /= m--;return ans % M;
}

方案2:

打表，C(n,m)=C(n-1,m-1)+C(n-1,m)，n<=10,000

const int M = 10007; 
const int MAXN = 1000; 
int C[MAXN+1][MAXN+1]; 
void Initial() 
{ int i,j; for(i=0; i<=MAXN; ++i) { C[0][i] = 0; C[i][0] = 1; } for(i=1; i<=MAXN; ++i) { for(j=1; j<=MAXN; ++j) C[i][j] = (C[i-1][j] + C[i-1][j-1]) % M; } 
} 
int Combination(int n, int m) 
{ return C[n][m]; 
} const int M = 10007;
const int MAXN = 1000;
int C[MAXN+1][MAXN+1];
void Initial()
{int i,j;for(i=0; i<=MAXN; ++i){C[0][i] = 0;C[i][0] = 1;}for(i=1; i<=MAXN; ++i){for(j=1; j<=MAXN; ++j)C[i][j] = (C[i-1][j] + C[i-1][j-1]) % M;}
}
int Combination(int n, int m)
{return C[n][m];
}

方案3:

质因数分解，C(n,m)=n!/(m!*(n-m)!)，C(n,m)=p1a1-b1-c1p2a2-b2-c2…pkak-bk-ck,n<=10,000,000

#include <cstdio>  
const int maxn=1000000; 
#include <vector>  
using namespace std; 
bool arr[maxn+1]={false}; 
vector<int> produce_prim_number() 
{ vector<int> prim; prim.push_back(2); int i,j; for(i=3;i*i<=maxn;i+=2) { if(!arr[i]) { prim.push_back(i); for(j=i*i;j<=maxn;j+=i) arr[j]=true; } } while(i<maxn) { if(!arr[i]) prim.push_back(i); i+=2; } return prim; 
} 
//计算n!中素数因子p的指数  
int cal(int x,int p) 
{ int ans=0; long long rec=p; while(x>=rec) { ans+=x/rec; rec*=p; } return ans; 
} 
//计算n的k次方对m取模，二分法  
int pow(long long n,int k,int M) 
{ long long ans=1; while(k) { if(k&1) { ans=(ans*n)%M; } n=(n*n)%M; k>>=1; } return ans; 
} 
//计算C（n，m）  
int combination(int n,int m) 
{ const int M=10007; vector<int> prim=produce_prim_number(); long long ans=1; int num; for(int i=0;i<prim.size()&&prim[i]<=n;++i) { num=cal(n,prim[i])-cal(m,prim[i])-cal(n-m,prim[i]); ans=(ans*pow(prim[i],num,M))%M; } return ans; 
} 
int main() 
{ int m,n; while(~scanf("%d%d",&m,&n),m&&n) { printf("%d\n",combination(m,n)); } return 0; 
} 

方案4:

Lucas定理，将m,n化为p进制,有:C(n,m)=C(n0,m0)*C(n1,m1)...(mod p)，算一个不是很大的C(n,m)%p,p为素数，化为线性同余方程,用扩展的欧几里德定理求解，n在int范围内，修改一下可以满足long long范围内。

#include <stdio.h>
const int M = 2013;
int ff[M+5];  //打表，记录n!，避免重复计算//求最大公因数
int gcd(int a,int b)
{if(b==0)return a;elsereturn gcd(b,a%b);
}//解线性同余方程，扩展欧几里德定理
int x,y;
void Extended_gcd(int a,int b)
{if(b==0){x=1;y=0;}else{Extended_gcd(b,a%b);long t=x;x=y;y=t-(a/b)*y;}
}//计算不大的C(n,m)
int C(int a,int b)
{if(b>a)return 0;b=(ff[a-b]*ff[b])%M;a=ff[a];int c=gcd(a,b);a/=c;b/=c;Extended_gcd(b,M);x=(x+M)%M;x=(x*a)%M;return x;
}//Lucas定理
int Combination(int n, int m)
{int ans=1;int a,b;while(m||n){a=n%M;b=m%M;n/=M;m/=M;ans=(ans*C(a,b))%M;}return ans;
}int main()
{int i,m,n;ff[0]=1;for(i=1; i<=M; i++) //预计算n!ff[i]=(ff[i-1]*i)%M;while(~scanf("%d%d",&n, &m)){printf("%d\n",Combination(n,m));}return 0;
}