一般的排列组合计数公式
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.
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.
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 2
31.
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.
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
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;
}