POJ 1650 Integer Approximation

Integer Approximation

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5081		Accepted: 1652

Description

The FORTH programming language does not support floating-point arithmetic at all. Its author, Chuck Moore, maintains that floating-point calculations are too slow and most of the time can be emulated by integers with proper scaling. For example, to calculate the area of the circle with the radius R he suggests to use formula like R * R * 355 / 113, which is in fact surprisingly accurate. The value of 355 / 113 ≈ 3.141593 is approximating the value of PI with the absolute error of only about 2*10 ^-7. You are to find the best integer approximation of a given floating-point number A within a given integer limit L. That is, to find such two integers N and D (1 <= N, D <= L) that the value of absolute error |A - N / D| is minimal.

Input

The first line of input contains a floating-point number A (0.1 <= A < 10) with the precision of up to 15 decimal digits. The second line contains the integer limit L. (1 <= L <= 100000).

Output

Output file must contain two integers, N and D, separated by space.

Sample Input

3.14159265358979
10000

Sample Output

355 113
题目大意：给出一个数字x和一个范围，在这个方位内招两个数字，是它们的商最接近x。
解题方法：直接枚举，取a,b为1每次对a,b求商，如果a / b > x，则a增加1，否则b增加1，每次记录下差值最小时a,b的值。

#include <stdio.h>

int main()
{
    double x, a, b, n, Min, n1, n2;
    scanf("%lf%lf", &x, &n);
    a = 1;
    b = 1;
    Min = n + 1;
    while(a <= n && b <= n)
    {
        if (a / b > x)
        {
            if (a / b - x < Min)
            {
                Min = a / b - x;
                n1 = a;
                n2 = b;
            }
            b++;
        }
        else
        {
            if (x - a / b < Min)
            {
                Min = x - a / b;
                n1 = a;
                n2 = b;
            }
            a++;
        }
    }
    printf("%.0f %.0f\n", n1, n2);
    return 0;
}