[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;
}