Uva--10375（数论，组合，唯一分解）

最新推荐文章于 2021-09-11 22:01:28 发布

转载最新推荐文章于 2021-09-11 22:01:28 发布 · 136 阅读

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原文链接：http://www.cnblogs.com/naturepengchen/articles/3949945.html

本文介绍了一种计算两个组合数比值的方法，通过唯一素数分解的方式，避免了直接计算阶乘带来的数值溢出问题。适用于p, q, r, s等数值在10000以内的场景。

2014-09-01 20:08:45

Problem D: Choose and divide

The binomial coefficient C(m,n) is defined as

         m!
C(m,n) = --------
         n!(m-n)!

Given four natural numbers p, q, r, and s, compute the the result of dividing C(p,q) by C(r,s).

The Input

Input consists of a sequence of lines. Each line contains four non-negative integer numbers giving values for p,q, r, and s, respectively, separated by a single space. All the numbers will be smaller than 10,000 with p>=q andr>=s.

The Output

For each line of input, print a single line containing a real number with 5 digits of precision in the fraction, giving the number as described above. You may assume the result is not greater than 100,000,000.

Sample Input

10 5 14 9
93 45 84 59
145 95 143 92
995 487 996 488
2000 1000 1999 999
9998 4999 9996 4998

Output for Sample Input

0.12587
505606.46055
1.28223
0.48996
2.00000
3.99960

思路：第二版小白书例题，参考了书里的思路，C(m,n) = m!/n!/(m-n)!，所以C(p,q) / C(r,s) = (p!*s!*(r-s)!) / (q!*r!*(p-q)!)，因为p，q，r，s的大小都在10000以内，所以考虑分别对分子和分母进行唯一素数分解，从而求出答案的唯一素数分解。

 1 /*************************************************************************
 2     > File Name: 10375.cpp
 3     > Author: Nature
 4     > Mail: 564374850@qq.com 
 5     > Created Time: Mon 01 Sep 2014 07:25:01 PM CST
 6 ************************************************************************/
 7 
 8 #include <cstdio>
 9 #include <cstring>
10 #include <cstdlib>
11 #include <cmath>
12 #include <vector>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16 
17 int anti_prime[10005];
18 int e[10005];
19 int p,q,r,s;
20 double ans;
21 vector<int> Prime;
22 
23 void Erato(int n){
24     memset(anti_prime,0,sizeof(anti_prime));
25     int m = sqrt(n + 0.5); //limit
26     for(int i = 2; i <= m; ++i) if(!anti_prime[i])
27         for(int j = i * i; j <= n; j += i) anti_prime[j] = 1;
28 }
29 
30 void Cal_factorial(int n,int tag){
31     for(int i = 2; i <= n; ++i){
32         int v = i;
33         for(int j = 0; j < Prime.size(); ++j){
34             while(v % Prime[j] == 0){
35                 v /= Prime[j];
36                 e[j] += tag;
37             }
38             if(v == 1) break;
39         }
40     }
41 }
42 
43 int main(){
44     Erato(10000);
45     for(int i = 2; i <= 10000; ++i) if(!anti_prime[i])
46         Prime.push_back(i);
47     while(scanf("%d%d%d%d",&p,&q,&r,&s) != EOF){
48         ans = 1.0;
49         memset(e,0,sizeof(e));
50         Cal_factorial(p,1);
51         Cal_factorial(s,1);
52         Cal_factorial(r - s,1);
53         Cal_factorial(q,-1);
54         Cal_factorial(p - q,-1);
55         Cal_factorial(r,-1);
56         for(int i = 0; i < Prime.size(); ++i)
57             ans *= pow(1.0 * Prime[i],1.0 * e[i]);
58         printf("%.5lf\n",ans);
59     }
60     return 0;
61 }
62 
63 
64

转载于:https://www.cnblogs.com/naturepengchen/articles/3949945.html