POJ 1306 Combinations 高精度乘法

题目大意：给出mn，让你求C(m,n)。

思路：公式都给你了，就100，暴力就可以关键还是高精度。如果按照算法“它让你怎么做你就怎么做”，那么很显然你需要写一个高精度除法。然而可以证明，这个除法是不会产生余数的。所以我们可以数论分析，然后避免高精度除法。

方法就是暴力求每个数的质因数，然后把被除数和除数相同的质因数消去，最后除数肯定会被消没。这样只要做高精度乘法就可以了。

CODE：

#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define MAX 110
#define BASE 1000
using namespace std;

int n,m;
int factor[MAX];

struct BigInt{
	int num[MAX],len;

	BigInt(int _ = 0) {
		memset(num,0,sizeof(num));
		if(_) {
			len = 1;
			num[1] = _;
		}
		else	len = 0;
	}
	BigInt operator *(int a) {
		BigInt re;
		int temp = 0;
		for(int i = 1; i <= len; ++i) {
			re.num[i] = num[i] * a + temp;
			temp = re.num[i] / BASE;
			re.num[i] %= BASE;
		}
		re.len = len;
		while(temp) {
			re.num[++re.len] = temp % BASE;
			temp /= BASE;
		}
		return re;
	}
	void operator *=(int a) {
		*this = *this * a;
	}
};

ostream &operator <<(ostream &os,const BigInt &a)
{
	os << a.num[a.len];
	for(int i = a.len - 1; i; --i)
		os << fixed << setfill('0') << setw(3) << a.num[i];
	return os;
}

inline void Divide(int x,int c)
{
	for(int i = 2; i * i <= x; ++i)
		if(x % i == 0)
			while(x % i == 0) {
				factor[i] += c;
				x /= i;
			}
	if(x != 1)	factor[x] += c;
}

int main()
{
	while(scanf("%d%d",&n,&m),m + n) {
		memset(factor,0,sizeof(factor));
		for(int i = 1; i <= n; ++i)
			Divide(i,1);
		for(int i = 1; i <= m; ++i)
			Divide(i,-1);
		for(int i = 1; i <= n - m; ++i)
			Divide(i,-1);
		BigInt ans(1);
		for(int i = 1;i <= 100; ++i)
			for(int j = 1; j <= factor[i]; ++j)
				ans *= i;
		cout << n << " things taken " << m << " at a time is " << ans <<" exactly." << endl;
	}
	return 0;
}