POJ 1306 Combinations

Combinations

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8013		Accepted: 3746

Description

Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following: 
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N 
Compute the EXACT value of: C = N! / (N-M)!M! 
You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long. For the record, the exact value of 100! is: 
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000 

Input

The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read.

Output

The output from this program should be in the form: 
N things taken M at a time is C exactly. 

Sample Input

100  6
20  5
18  6
0  0

Sample Output

100 things taken 6 at a time is 1192052400 exactly.
20 things taken 5 at a time is 15504 exactly.
18 things taken 6 at a time is 18564 exactly.

题意概述：其实就是求排列组合。

解题思路：30！已经超出了整数所能表示的范围，若要使用C或c++，只有用字符串模拟，但这样的话，不仅复杂度急剧增大，而且很可能会超时；此时，java处理大数据是不二选择。

源代码：

import java.math.BigInteger;
import java.util.Scanner;


public class Main {

	public static void main(String[] args) {
		
		Scanner cin=new Scanner(System.in);
		while(true){
			int a=cin.nextInt();
			int b=cin.nextInt();
			if(a==0 && b==0)
				return;
			BigInteger fir=F(a);
			BigInteger sec=F(a-b);
			BigInteger third=F(b);
			BigInteger result=fir.divide(sec).divide(third);
			System.out.println(a+" things taken "+b+" at a time is "+result.toString()+" exactly.");
		}
	}
	
	public static BigInteger F(int n){
		BigInteger result=BigInteger.ONE;//将result赋值为1
		for(int i=n;i>=2;i--){
			result=result.multiply(new BigInteger(String.valueOf(i)) );
		}
		return result;
	}


}