标签：ber 思路 each stream when output value leading air
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.
解题思路：数据比较小，直接暴力求解，类型全开long long即可，水过！
AC代码：
1 #include<iostream>
2 using namespace std;
3 typedef long long LL;
4 LL n,k,m,ans;
5 int main(){
6 while(cin>>n>>k&&(n+k)){
7 m=k;
8 if(n-k<k)k=n-k;
9 ans=1;
10 for(LL i=1;i<=k;++i)ans=ans*(n-i+1)/i;
11 cout<<n<<" things taken "<<m<<" at a time is "<<ans<<" exactly."<<endl;
12 }
13 return 0;
14 }
O - Combinations （组合数学）
标签：ber 思路 each stream when output value leading air
原文地址：https://www.cnblogs.com/acgoto/p/9315866.html
