VUA-10375 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：
计算组合数公式。
数据会很大，直接算会超范围。
观察一下，这个公式中分子和分母中相乘的数字个数是一样的。
所以可以乘一个数字的同时除一个数字，正好计算完毕，不会超范围
对于这个公式，进行化简，可以化简称两种情况，对应两种情况。
对r、s和p、q但读进行约分化简
p！/[q!(p-q)!] = p*(p-1)…(q+1)/(p-q)!
p！/[q!(p-q)!] =p*(p-1)…(p-q+1))/q!
和
s!(r-s)!/r! =(r-s)!/[r(r-1)…(s+1)]
s!(r-s)!/r! =s!/[r(r-1)…(r-s+1)]

这两种情况，根据p-q和q的大小（r-s和s的大小）来取舍，
显然选择计算量较小的。

#include <stdio.h>  int p, q, r, s, i;  double ans;  int main () {      while (~scanf("%d%d%d%d", &p, &q, &r, &s)) {          ans = 1.0;          if (p - q < q)              q = p - q;          if (r - s < s)              s = r - s;          for (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("%.5lf\n", ans);      }      return 0;  }