# POJ 1650 Integer Approximation C++版

POJ题解 1701浏览

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

Slyar:求出在L范围内的两个数，使得它们相除后减去A的结果的绝对值最小。暴力枚举就好了，这题很恶心…

```#include
#include

using namespace std;

int main()
{
double a, min, temp;
int i, j, l, n, d;

cin >> a >> l;
n = d = 1;
min = 20;

for (i = 1; i <= l; i++)
{
j = (int)(i / a);

if (j > l) continue;
temp = fabs(a - (double)i / (double)j);
if (temp < min)
{
min = temp;
n = i;
d = j;
}

j++;
if (j > l) continue;
temp = fabs(a - (double)i / (double)j);
if (temp < min)
{
min = temp;
n = i;
d = j;
}
}

cout << n << " " << d << endl;

//system("pause");
return 0;
}
```