Choose and divide

The binomial coefficient C(m, n) is defined as C(m, n) = m! (m − n)! n! Given four natural numbers p, q, r, and s, compute the the result of dividing C(p, q) by C(r, s). 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 and r ≥ s. 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

Sample Outpu

t 0.12587 505606.46055 1.28223 0.48996 2.00000 3.99960

 

唯一分解定理应用

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 10009
#define L 31
#define INF 1000000009
#define eps 0.00000001
/*
唯一分解定理 应用
*/
int prime[MAXN], e[MAXN];
void getprime()
{
    memset(prime, false, sizeof(prime));
    for (int i = 2; i <= MAXN; i++)
    {
        if (!prime[i])
            prime[++prime[0]] = i;
        for (int j = 1; j <= prime[0] && prime[j] <= MAXN / i; j++)
        {
            prime[prime[j] * i] = 1;
            if (i%prime[j] == 0) break;
        }
    }
}
void add_interger(int n, int d)
{
    for (int i = 1; i <= prime[0]; i++)
    {
        while (n%prime[i] == 0)
        {
            n /= prime[i];
            e[i] += d;
        }
        if (n == 1) break;
    }
}
void add_factorial(int n, int d)
{
    for (int i = 1; i <= n; i++)
        add_interger(i, d);
}
int main()
{
    getprime();
    int p, q, r, s;
    while (cin >> p >> q >> r >> s)
    {
        memset(e, 0, sizeof(e));
        add_factorial(p, 1);
        add_factorial(q, -1);
        add_factorial(p - q, -1);
        add_factorial(r, -1);
        add_factorial(s, 1);
        add_factorial(r - s, 1);
        double ans = 1;
        for (int i = 1; i <= prime[0]; i++)
            ans *= pow(prime[i], e[i]);
        printf("%.5lf\n", ans);
    }
    return 0;
}