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质因数分解：
Description  
Problem D: Choose and divide The binomial coefficient m!
C(m,n) = --------
n!(m-n)!
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 The OutputFor 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 Input10 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 Input0.12587 505606.46055 1.28223 0.48996 2.00000 3.99960 Source
Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths ::
Examples Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 6. Mathematical Concepts and Methods Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Mathematics :: Combinatorics :: Binomial Coefficients Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 5. Mathematics :: |
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#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=10010;
int p,q,r,s;
int prime[maxn],pn;
long long int fnum[maxn],pnum[maxn];
bool vis[maxn];
void pre_init()
{
memset(vis,true,sizeof(vis));
for(int i=2; i<maxn; i++)
{
if(i%2==0&&i!=2) continue;
if(vis[i]==true)
{
prime[pn++]=i;
for(int j=2*i; j<maxn; j+=i)
{
vis[j]=false;
}
}
}
}
void fenjie_x(int x,long long int* arr)
{
for(int i=0; i<pn&&x!=1; i++)
{
while(x%prime[i]==0)
{
arr[i]++;
x/=prime[i];
}
}
}
void fenjie(int x,long long int* arr)
{
for(int i=2; i<=x; i++)
fenjie_x(i,arr);
}
void jianshao()
{
for(int i=0; i<pn; i++)
{
long long int Min=min(fnum[i],pnum[i]);
fnum[i]-=Min;
pnum[i]-=Min;
}
}
int main()
{
pre_init();
while(scanf("%d%d%d%d",&p,&q,&r,&s)!=EOF)
{
memset(pnum,0,sizeof(pnum));
memset(fnum,0,sizeof(fnum));
fenjie(p,pnum);fenjie(s,pnum);fenjie(r-s,pnum);
fenjie(q,fnum);fenjie(r,fnum);fenjie(p-q,fnum);
jianshao();
double ans=1.;
for(int i=0; i<pn; i++)
{
while(pnum[i]--)
{
ans*=1.*prime[i];
}
while(fnum[i]--)
{
ans/=1.*prime[i];
}
}
printf("%.5lf\n",ans);
}
return 0;
}
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