[ACM] poj 2249 Binomial Showdown （排列组合公式优化）

Description
In how many ways can you choose k elements out of n elements, not taking order into account?

Write a program to compute this number.
Input
The input will contain one or more test cases.

Each test case consists of one line containing two integers n (n>=1) and k (0<=k<=n).

Input is terminated by two zeroes for n and k.
Output
For each test case, print one line containing the required number. This number will always fit into an integer, i.e. it will be less than 231.

Warning: Don't underestimate the problem. The result will fit into an integer - but if all intermediate results arising during the computation will also fit into an integer depends on your algorithm. The test cases will go to the limit.

Sample Input

4 2
10 5
49 6
0 0

Sample Output

6
252
13983816

Source
Ulm Local 1997

解题思路：
本题题意很明确，就是让求C（n,m）是多少，但传统的组合公式计算过程中会越界，所以采用另一种思路，把组合公式的分子和分母的数字分别存到两个数组中，然后两层循环，分别求出分子和分母的最大公约数，然后约分，这样把分子分母化成最简。最后再采用传统的组合公式就可以了。

代码：

#include <iostream>
using namespace std;
const int maxn=3000;
int up[maxn],down[maxn];//分别存分子和分母的数

int gcd(int a,int b)
{
if(!a)
return b;
int c;
while(b)
{
c=b;
b=a%b;
a=c;
}
return a;
}

int main()
{
int n,m;
while(cin>>n>>m&&(m||n))
{
if(m>n-m)
m=n-m;
int temp=n;
for(int i=1;i<=m;i++)//根据组合公式把原始的分子分母分别存在数组中
{
up[i]=temp--;
down[i]=i;
}
for(int i=1;i<=m;i++)//外层循环代表分母，对分母依次进行约分
for(int j=1;j<=m;j++)
{
temp=gcd(down[i],up[j]);
if(temp>1)
{
up[j]/=temp;
down[i]/=temp;
}
if(down[i]==1)//分母已经为1，退出，进行下一个分母的约分
break;
}
int mul=1;
for(int i=1;i<=m;i++)
mul=mul*up[i]/down[i];
cout<<mul<<endl;
}
return 0;
}