Combinations
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000Input
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 0Sample 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:[email protected]
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", n, m, ans);
58
59 return true;
60 }
61
62
63 int main() {
64 while (Do());
65 return 0;
66 }