UVA 10375 (13.11.07)

Problem D: Choose and divide

The binomial coefficient C(m,n) is defined as

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 Input

Input consists of a sequence of lines. Each line contains four non-negativeinteger 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 Output

For each line of input, print a single line containing a real numberwith 5 digits of precision in the fraction, giving the number asdescribed above. You may assume the result is not greater than100,000,000.

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

Output for Sample Input

0.12587

505606.46055

1.28223

0.48996

2.00000

3.99960

题意我就不说了~

错误思路; 首先我用了先全部乘起来后一起除的做法, 结果超出数据范围~, 所以这条路走不通了

正确做法: 因为这个式子分子分母的项数一样, 我们可采取乘一次除一次的方式, 不仅如此, 两个函数还能一起处理~

AC代码:

#include<stdio.h>  
  
double getSum(double m, double n, double x, double y) {  
    double sum = 1;  
    double l1, l2;  
    if(m - n < n)  
        l1 = m - n;  
    else  
        l1 = n;  
    if(x - y < y)  
        l2 = x - y;  
    else  
        l2 = y;  
    for(int i = 1; i <= l1 || i <= l2; i++) {  
        if(i <= l1)  
            sum = sum * (m - l1 + i) / i;  
        if(i <= l2)  
            sum = sum * i / (x - l2 + i);  
    }  
    return sum;  
}  
  
int main() {  
    double p, q, r, s;  
    while(scanf("%lf %lf %lf %lf", &p, &q, &r, &s) != EOF)  
        printf("%.5lf\n", getSum(p, q, r, s));  
    return 0;  
}

补充：软件开发 , C++ ,