POJ-1306(组合数) Combinations_poj - 1306 - combinations

本文链接：https://blog.youkuaiyun.com/MUHAHADAWANG/article/details/79674306

本文介绍了一个计算特定范围内组合数的程序实现。面对较大数值时，采用优化策略减少计算量，并提供了一个具体的C/C++实现案例。

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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.
1. 根据给的公式暴力求…

#include <iostream>
#include <cstdio>
using namespace std;
int n,m,t1,t2;
double ans1,ans2;
void solve(){
    t1 = n;
    t2 = m;
    if(n - m < m){  //根据c(n,n-1) == c(n,1)而来，可以优化时间
        m = n - m;
    }
    ans1 = 1.0; // ->   m!
    ans2 = 1.0; // ->   n!/(n-m)!
    for (int i = 1;i <= m;i++){
        ans1 *= i;
        ans2 *= n;
        n--;
    }
    printf("%d things taken %d at a time is %I64d exactly.\n",t1,t2,(ans2 / ans1));
}
int main(){
    while(~scanf("%d %d",&n,&m)){
        if (n==0&&m==0){
            break;
        }
        solve();
    }
    return 0;
}