唯一分解定理 （Choose and Divide Uva10375）

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 Output
0.12587
505606.46055
1.28223
0.48996
2.00000
3.99960

简单翻译一下，告诉你p,q,r,s求C(p,q)/C(r,s)

简单的暴力是不可能过得，所以我们要用数学的方法，我们运用唯一分解法，将质因数分解之后存在数组中，这样乘除就是这个因数的个数的加减，最后将所有因数乘起来就可以了~

#include<bits/stdc++.h>using namespace std;vector<int> prime;int book[20000],e[20000];void add_integer(int n,int d){for(int i=0;i<prime.size();i++){while(n%prime[i]==0){e[i]+=d;n/=prime[i];}if(n==1) break;}}void add(int n,int d){for(int i=1;i<=n;i++)add_integer(i,d);}void init(){book[1]=1;for(int i=2;i<=101;i++)if(!book[i])for(int j=i*i;j<=10000;j+=i)book[j]=1;for(int i=1;i<=10000;i++)if(!book[i])prime.push_back(i);}double Pow(int a,int b){int ok=0;if(b==0) return 1;if(b<0) ok=1,b=-b;double ans=1;for(int i=1;i<=b;i++)ans*=double(a);if(ok) return double(1/ans);return ans;}int main(){init();int p,q,r,s;while(cin>>p>>q>>r>>s){memset(e,0,sizeof(e));add(p,1);add(q,-1);add(p-q,-1);add(r,-1);add(s,1);//作为除数的话和前面的正好相反 add(r-s,1);double ans=1;for(int i=0;i<prime.size();i++)ans*=Pow(prime[i],e[i]);printf("%.5lf\n",ans);}return 0;}