求組合數----模闆

模數為素數時，根據費馬小定理論用快速幂求 

#include <stdio.h>
#include <iostream>
using namespace std;
const int MOD=998244353;
const int MAXN=202020;
int fac[MAXN],facinv[MAXN];
 
long long quickmul(int a,int b)
{
	long long ret=1;
	for(; b ; b >>=1 ,a =(long long) a * a % MOD)
		if((b & 1))
			ret=ret * a % MOD;
	return ret;
}
long long C(int n,int m)
{
	if(n<0||m<0||n<m)
		return 0;
	return (long long)fac[n]*facinv[m]%MOD*facinv[n-m]%MOD;
} 
void init()
{
	fac[0]=1;
	for(int i=1;i<=MAXN;i++)
		fac[i]=(long long)fac[i-1]*i%MOD;
	facinv[MAXN]=quickmul(fac[MAXN],MOD-2);
	for(int i=MAXN;i>0;i--)
		facinv[i-1]=(long long)facinv[i]*(i)%MOD;
		//1/(i-1)!=i/(i)!
}
int main()
{
	init();
	int n,m;
	while(1)
	{
		scanf("%d %d",&n,&m);
		printf("%lld\n",C(n,m));
	}
}
           

模數不為逆元，根據楊輝三角

#include <bits/stdc++.h>
using namespace std;
int C[2000+5][2000+5];
void getC()
{
	for( int i=1; i<=2000; i++ )
	{
		C[i][0] = C[i][i] = 1;	//初始化第一列和對角線位置全為1，即最兩邊的位置 
		for( int j=1; j<i; j++ )
			C[i][j] = (C[i-1][j] + C[i-1][j-1]) % 1007;
	}
}
int main()
{
	int m, n;
    getC(); 
    while(scanf("%d %d",&n,&m))  
    {  
        printf("%d\n", C[n][m] ); //第n行第m列即為從n中選m個 
    }  
	return 0;
}