poj 2249 Binomial Showdown(组合数 公式优化)

//  组合数学,开始了……

题目地址 : poj 2249 Binomial Showdown

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

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组合数公式的优化

普通的组合数公式：C(n,m) = n!/((n-m)!*m!) = (n * n-1 * n-2 * n-3 * n-4 * n-5 * n-6 * …… * n-m) / (m * m-1 * m-2 * m-3 * m-4 * m-5 * …… * 2 * 1)

优化 ：将分子和分母 的因子分别存到一个数组中 nn[]  (分子)  mm[] (分母),双重循环遍历, 进行各个因子约分，因为 组合数为一个整数 ，即N % M = 0,

所以mm[] 的元素一定可以全部约分为1，然后只需将 mm[] 中的元素相乘就行了。。

******************************/#include <iostream>
#include<string.h>
#include <stdio.h>
using namespace std;
const int N = 10000;
int nn
,mm
;
int gcd(int a,int b)// 求n,m 的最大公约数
{
return (b==0)?a:gcd(b,a%b);
}
int main()
{
int i,j,n,m,t,h,sum,temp;
while(cin>>n>>m&&(n||m))// 程序结束条件是 n,m 中一个为0就行，用||,我开始用&& WA了好多次，要注意
{
t = h = 0;
sum = 1;
//cout<<gcd(m,n)<<endl;
if(m>n-m)
m = n-m;
for(i = n-m+1;i<=n;i++)// 分子赋值
nn[t++] = i;
for(i = 1;i<=m;i++)// 分母赋值
mm[h++] = i;
for(i = 0;i<m;i++)
{
for(j = 0;j<m;j++)
{
temp = gcd(nn[i],mm[j]);// 约分
nn[i] = nn[i]/temp;
mm[j] = mm[j]/temp;
}
}
for(i = 0;i<m;i++)
sum*=nn[i];
cout<<sum<<endl;
}
}