UVa10375

Choose and divide
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

题意：

       设C(m,n) = m! / ((m - n)! * n!)，输入正整数p、q、r、s，输出C(p,q)/C(r,s)

 

分析：

       C(p,q) / C(r,s) = p! * (r - s)! * s! / ((p - q)! * q! * r!)，于是，我们采用唯一分解定理。add_integer(n,d)表示将整数n进行因数分解。add_factorial(n, d)表示将n!进行因数分解。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<iostream>
using namespace std;
const int maxn = 10000;
vector<int> primes;
int e[maxn];
// 乘以或除以n. d=0表示乘，d=-1表示除
void add_integer(int n, int d) {
  for(int i = 0; i < primes.size(); i++) {
    while(n % primes[i] == 0) {
      n /= primes[i];
      e[i] += d;
    }
    if(n == 1) break; // 提前终止循环，节约时间
  }
}

void add_factorial(int n, int d) {
  for(int i = 1; i <= n; i++)
    add_integer(i, d);
}

bool is_prime(int n) {
  int m = floor(sqrt(n) + 0.5);
  for(int a = 2; a <= m; a++)
    if(n % a == 0) return false;
  return true;
}
int main() {
  for(int i = 2; i <= 10000; i++)
    if(is_prime(i)) primes.push_back(i);
  int p, q, r, s;
  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;
    for(int i = 0; i < primes.size(); i++)
      ans *= pow(primes[i], e[i]);
    printf("%.5lf\n", ans);
  }
  return 0;
}