质因数分解:
Choose and divide
Description Problem D: Choose and divideThe 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 InputInput 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 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 Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: Mathematics :: Combinatorics :: Binomial Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 5. Mathematics :: Combinatorics Root :: Prominent Problemsetters :: Gordon V. Cormack |
#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; }