UVA 10375 Choose and divide 計算組合數C(p,q)/C(r,s)。

題目鏈接http://www.bnuoj.com/v3/problem_show.php?pid=18793

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 and r>=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

下面給出兩種解法：

1.

分解每個乘數的質因子，然後利用 質因子的表示方法 去連乘，比如50=(2^1)*(5^2)，可參考紫書317頁

//分解每個乘數的質因子，然後利用 質因子的表示方法 去連乘，比如50=(2^1)*(5^2)
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
const int MAX=10005;
int p,q,r,s,e[MAX]; //e存儲因子的個數
void add_factorial(int n,int d) //分解質因子
{
    for(int i=2;i<MAX;i++){
        while(n%i==0){
            n/=i;
            e[i]+=d;
        }
        if(n==1) break;
    }
}
void solve()
{
    memset(e,0,sizeof(e));

    //...C(p,q)
    for(int i=p;i>=p-q+1;i--) add_factorial(i,1);
    for(int i=q;i>=1;i--) add_factorial(i,-1);

    //...1/C(r,s)
    for(int i=r;i>=r-s+1;i--) add_factorial(i,-1);
    for(int i=s;i>=1;i--) add_factorial(i,1);

    double ans=1;
    for(int i=2;i<MAX;i++)
        ans*=pow(i,e[i]);
    printf("%.5lf\n",ans);
}
int main()
{
    while(~scanf("%d%d%d%d",&p,&q,&r,&s))
    {
        solve();
    }
    return 0;
}

2.

//將組合數化簡一下邊乘邊除
#include<stdio.h>
int p,q,r,s;
void solve()
{
    double ans = 1.0;
    if(p-q<q) q=p-q; //因爲C(p,q)=C(p,p-q)
    if(r-s<s) s=r-s; //同理
    for(int i=1;i<=p||i<=r;i++){
        if(i<=q) ans=ans*(p-i+1)/i;
        if(i<=s) ans=ans/(r-i+1)*i;
    }
    printf("%.5lf\n", ans);
}
int main()
{
    while(scanf("%d%d%d%d",&p,&q,&r,&s)==4)
        solve();
    return 0;
}