Uva--10375（数论，组合，唯一分解）

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。

/*************************************************************************
    > File Name: 10375.cpp
    > Author: Nature
    > Mail: 564374850@qq.com 
    > Created Time: Mon 01 Sep 2014 07:25:01 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int anti_prime[10005];
int e[10005];
int p,q,r,s;
double ans;
vector<int> Prime;

void Erato(int n){
    memset(anti_prime,0,sizeof(anti_prime));
    int m = sqrt(n + 0.5); //limit
    for(int i = 2; i <= m; ++i) if(!anti_prime[i])
        for(int j = i * i; j <= n; j += i) anti_prime[j] = 1;
}

void Cal_factorial(int n,int tag){
    for(int i = 2; i <= n; ++i){
        int v = i;
        for(int j = 0; j < Prime.size(); ++j){
            while(v % Prime[j] == 0){
                v /= Prime[j];
                e[j] += tag;
            }
            if(v == 1) break;
        }
    }
}

int main(){
    Erato(10000);
    for(int i = 2; i <= 10000; ++i) if(!anti_prime[i])
        Prime.push_back(i);
    while(scanf("%d%d%d%d",&p,&q,&r,&s) != EOF){
        ans = 1.0;
        memset(e,0,sizeof(e));
        Cal_factorial(p,1);
        Cal_factorial(s,1);
        Cal_factorial(r - s,1);
        Cal_factorial(q,-1);
        Cal_factorial(p - q,-1);
        Cal_factorial(r,-1);
        for(int i = 0; i < Prime.size(); ++i)
            ans *= pow(1.0 * Prime[i],1.0 * e[i]);
        printf("%.5lf\n",ans);
    }
    return 0;
}