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 dividingC(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 forp, q, r, and s, respectively, separated by a single space. All the numbers will be smaller than 10,000 withp>=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 14 993 45 84 59145 95 143 92995 487 996 4882000 1000 1999 9999998 4999 9996 4998

Output for Sample Input

0.12587505606.460551.282230.489962.000003.99960

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;}