POJ 1306.Combinations

Combinations
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
SubmitStatus Practice POJ 1306
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

 1 /*
 2 By:OhYee
 3 Github:OhYee
 4 Email:oyohyee@oyohyee.com
 5 Blog:http://www.cnblogs.com/ohyee/
 6 
 7 かしこいかわいい？
 8 エリーチカ！
 9 要写出来Хорошо的代码哦~
10 */
11 #include <cstdio>
12 #include <algorithm>
13 #include <cstring>
14 #include <cmath>
15 #include <string>
16 #include <iostream>
17 #include <vector>
18 #include <list>
19 #include <queue>
20 #include <stack>
21 using namespace std;
22 
23 //DEBUG MODE
24 #define debug 0
25 
26 //循环
27 #define REP(n) for(int o=0;o<n;o++)
28 
29 unsigned long long gcd(unsigned long long a, unsigned long long b) {
30     return b == 0 ? a : gcd(b, a%b);
31 }
32 
33 bool Do() {
34     int n, m;
35     if (scanf("%d%d", &n, &m), n == 0 && m == 0)
36         return false;
37 
38     unsigned long long ans = 1;
39     int a = max(m, n - m);
40     int b = min(m, n - m);
41     unsigned long long t = 1;
42     for (int i = n, j = 2; i > a; i--, j++) {
43         ans *= i;
44         if (j <= b || t > 1) {
45             if (j <= b)
46                 t *= j;
47             if (t > 1) {
48                 unsigned long long q = gcd(ans, t);
49                 ans /= q;
50                 t /= q;
51             }
52         }
53 
54     }
55 
56 
57     printf("%d things taken %d at a time is %llu exactly.
", n, m, ans);
58 
59     return true;
60 }
61 
62 
63 int main() {
64     while (Do());
65     return 0;
66 }