Combinations（组合计数）
Link：http://poj.org/problem?id=1306
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Combinations
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
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AC code：
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<vector>
#define LL long long
#define MAXN 1000100
using namespace std;
LL n,m;
LL Cnm(LL n,LL m)//计算组合数C(n,m)
{
if(m>n/2)//根据组合公式，可以减少枚举量
m=n-m;
LL a=1,b=1;
for(int i=1;i<=m;i++)//顺序进行m次运算
{
a*=n+1-i;//计算前i项运算结果的分子a和分母 b
b*=i;
if(a%b==0)
{
a/=b;
b=1;
}
}
return a/b;
}
int main()
{
while(cin>>n>>m)
{
if(n==0&&m==0)
break;
cout<<n<<" things taken "<<m<<" at a time is "<<Cnm(n,m)<<" exactly."<<endl;
}
return 0;
}