UVA 10375 Choose and divide（数论）

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

唯一分解定理：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
#include<vector>
#include<cmath>
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
const int maxn=10005;
int e[maxn];
int p,q,r,s;
int visit[maxn];
vector<int>prime;
void is_prime(int n)
{
    int m=sqrt(n+0.5);
    memset(visit,0,sizeof(visit));
    for(int i=2;i<=m;i++)
    {
        for(int j=i*i;j<=n;j+=i)
          visit[j]=1;
    }
}
void get_prime(int n)
{
    for(int i=2;i<=n;i++)
    {
        if(!visit[i])
            prime.push_back(i);
    }
}
void add_integer(int n,int d)
{
    for(int i=0;i<prime.size();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_integer(i,d);
}
int main()
{
    is_prime(maxn-1);
    get_prime(maxn-1);
    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=1.0;
        for(int i=0;i<prime.size();i++)
            ans*=pow(prime[i],e[i]);
        printf("%.5f\n",ans);
    }
    return 0;
}

还有一种方法：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
int p,q,r,s;

int main()
{
    while(~scanf("%d%d%d%d",&p,&q,&r,&s))
    {
        if(p-q<q)//q和p-q中较大值的阶乘和p的阶乘消去
           q=p-q;
        if(r-s<s)
           s=r-s;
        double ans=1.0;
        for(int i=1;i<=q||i<=s;i++)
        {
            if(i<=q)
                ans=ans*(p-q+i)/i;
            if(i<=s)
                ans=ans/(r-s+i)*i;
        }
        printf("%.5f\n",ans);
    }
    return 0;
}

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