TOJ 1138. Binomial Showdown

最新推荐文章于 2018-09-02 08:49:54 发布

原创最新推荐文章于 2018-09-02 08:49:54 发布 · 681 阅读

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#integer #sun #iostream #input #each #output

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本文介绍了一个计算组合数C(k,n)的程序实现方法，并通过示例输入输出展示了如何避免中间计算过程中可能出现的整数溢出问题。该方法适用于n较大但结果依然可以装入整型变量的情况。

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 Specification

The input file 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 Specification

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

这题很纠结，就是算C（k,n）,第一次函数是这样写的sun=sun*(n-i)/i;结果RE……应该是因为乘的过程中大头除小头超了，而sun=sun*(n-m+i)/i;则是小头除小头。

code:
#include <iostream> 
using namespace std;   
int main() 
{ 
	int n,m,i; 
	double sun;
	while(scanf("%d%d",&n ,&m ) != EOF ) 
	{ 
   		if(n==0&&m==0) break; 
   		if(m>n/2) m=n-m; 
   		sun=1; 
   		for(i=1;i<=m;i++) sun=sun*(n-m+i)/i;
   		printf("%lld/n", (long long)sun );    
	} 
 	return 0; 
}