poj_1306_Combinations

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.

#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
#define ll long long
#define M 105

ll n,k;

int main()
{
    ll i,j;
    while(cin>>n>>k)
    {
        if(n==0&&k==0)
            break;
        if(k==n)
        {
            cout<<n<<" things taken "<<k<<" at a time is "<<1<<" exactly."<<endl;
            continue;
        }
        cout<<n<<" things taken "<<k<<" at a time is ";
        if(n-k<k)
            k=n-k;
        ll ans=1;
        for(i=1;i<=k;i++)
        {
            ans=ans*(n-i+1)/i;
        }
        cout<<ans<<" exactly."<<endl;
    }
}

　　

转载于:https://www.cnblogs.com/ygtzds/p/8039988.html