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 在n个不同的元素中：

    若取 r 个按次序排列， 则成为从n中取r个排列，其排列数为：P( n, r )=( n! ) / ( n-r )!  。

    如取出 r 个二不考虑起次序，则称为 从n中取 r 个组合， 其组合数为：C( n, r ) = ( n! )/[ ( r! ) *( n-r )! ] 。

 

 进行阶乘运算的数值较大， 直接计算分子和分母的话，效率低切容易溢出。

   一 ： 采取：连乘 r 个整商法

       C( n, r ) = [ (n-r-1) / r ] * [ (n-r-2)/(r-1) ] * ..........* [ n / 1 ]  .

  二： 二项式洗漱公式法

      C( j, i ) =  C( j, i-1) + C(j-1, i-1 ) .

     通过 i, j 的双重循环直接递推 c[i][j] ;

  样题：

  POJ  2249

        因为常用blockcodes编译代码，所以对于大的数据类型会选择 long long,  对于这道题，用了__int64类型（因为不常用，所以不能出错）

       复习一下：定义 __int64 ( 双下划线+int+64 )   输出：printf("I64d\n", a);     I64d    !!!!

       

Binomial Showdown

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18112		Accepted: 5514

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

采用 分整式连乘的方法

C ( n, k ) = (n-k+1)/k  *   (n-k+2)/(k-1)  * ......  * ( n)/1  .

#include <stdio.h>
#include <string.h>
#include <stdlib.h>__int64 work(__int64 n, __int64 k)
{if(k>n/2)k = n-k; //减少枚举量（剪枝）
__int64 a=1, b=1; //代表分子和分母int i;for(i=1; i<=k; i++)  //循环k次运算{a=a*(n-i+1); //分子b=b*i;if(a%b==0) //说明能整除，进行整数商处理{a=a/b;b=1;}}return a/b;
}int main()
{int n, k;while(scanf("%d %d", &n, &k)!=EOF){if(n==0  ){break;}printf("%I64d\n", work(n, k) );}return 0;
}