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: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 }