POJ 1306.Combinations

410


题目




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.


题解



题目题意比较简单 计算N!/(N-M)!M!
关键在于数值的计算上
尽管最后结果我们或许可以保存下,但是其中间要乘到很大的数再除下去,因此要尽可能让中间数小

由于N>M,我们可以剩下很大一部分乘法,只需计算N*(N-1)*······*(M+2)*(M+1)

因此,比较下M和N-M,选择其中较大的与N!约分

然后在计算另一部分,分母和分子同时乘数,每乘一次进行一次约分(gcd)

这样就能在不溢出的情况下计算出我们想要的答案

代码



/*
By:OhYee
Github:OhYee
Email:oyohyee@oyohyee.com
Blog:http://www.cnblogs.com/ohyee/

かしこいかわいい?
エリーチカ!
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
using namespace std;

//DEBUG MODE
#define debug 0

//循环
#define REP(n) for(int o=0;o<n;o++)

unsigned long long gcd(unsigned long long a, unsigned long long b) {
    return b == 0 ? a : gcd(b, a%b);
}

bool Do() {
    int n, m;
    if (scanf("%d%d", &n, &m), n == 0 && m == 0)
        return false;

    unsigned long long ans = 1;
    int a = max(m, n - m);
    int b = min(m, n - m);
    unsigned long long t = 1;
    for (int i = n, j = 2; i > a; i--, j++) {
        ans *= i;
        if (j <= b || t > 1) {
            if (j <= b)
                t *= j;
            if (t > 1) {
                unsigned long long q = gcd(ans, t);
                ans /= q;
                t /= q;
            }
        }

    }


    printf("%d things taken %d at a time is %llu exactly.\n", n, m, ans);

    return true;
}


int main() {
    while (Do());
    return 0;
}
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