POJ 1306 Combinations 组合数

Combinations

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10205		Accepted: 4650

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.

Source

UVA Volume III 369

m'ban

模板题，求组合数汇总（点这里）

#include<cstdio>  
#include<cstring>  
#include<cstdlib>  
#include<cctype>  
#include<cmath>  
#include<iostream>  
#include<sstream>  
#include<iterator>  
#include<algorithm>  
#include<string>  
#include<vector>  
#include<set>  
#include<map>  
#include<stack>  
#include<deque>  
#include<queue>
using namespace std;
typedef long long ll;
ll a[100000000];
 
ll C(ll n,ll m)
{

		if (n - m < m)
			m = n - m;
		a[0] = 1;
		for (int i = 1; i <= m; i++) 
		{
			a[i] = (n - i + 1) * a[i - 1] / i;
		}
	return a[m];
}
int main()
{
	ll n,m;
	while (~scanf("%lld%lld",&n,&m) && (n || m))
	{
		
	printf("%lld things taken %lld at a time is %lld exactly.\n",n,m,C(n,m));
	} 
}