追赶法搜索

Integer Approximation

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5827		Accepted: 1968

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

POJ1650

最初这题是在算法竞赛宝典二分中看到的，大意是可以枚举分母二分分子。

挺麻烦的，后来查了下可以利用追赶法搜索。

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    double ans;
    int limit;
    while(cin>>ans>>limit)
    {
        int l=1,r=1;
        int ansl=1,ansr=1;
        double lastcha=999999;
        while(l<=limit&&r<=limit)
        {
            double chazhi=ans-(double)l/r;
            double chazhi2=fabs(chazhi);
            if(chazhi2<lastcha)
            {
                lastcha=chazhi2;/// 不要写成chazhi!!!
                ansl=l,ansr=r;
            }
            if(chazhi>0)
                l++;
            else
                r++;
        }
        cout<<ansl<<' '<<ansr<<endl;
    }
}

另外还有一题POJ3039

Close Encounter

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3547		Accepted: 1306

Description

Lacking even a fifth grade education, the cows are having trouble with a fraction problem from their textbook. Please help them. The problem is simple:

Given a properly reduced fraction (i.e., the greatest common divisor of the numerator and denominator is 1, so the fraction cannot be further reduced) find the smallest properly reduced fraction with numerator and denominator in the range 1..32,767 that is closest (but not equal) to the given fraction.

Input

* Line 1: Two positive space-separated integers N and D (1 <= N < D <= 32,767), respectively the numerator and denominator of the given fraction

Output

* Line 1: Two space-separated integers, respectively the numerator and denominator of the smallest, closest fraction different from the input fraction.

Sample Input

2 3

Sample Output

21845 32767

Hint

INPUT DETAILS:
2/3

OUTPUT DETAILS:
21845/32767 = .666676839503.... ~ 0.666666.... = 2/3.

同样原理

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    double ans;
    int a,b;
    int limit=32767;
    while(cin>>a>>b)
    {
        ans=(double)a/b;
        int l=1,r=1;
        int ansl=1,ansr=1;
        double lastcha=999999;
        while(l<=limit&&r<=limit)
        {
            double chazhi=ans-(double)l/r;
            double chazhi2=fabs(chazhi);
            if(chazhi2<lastcha&&!(l%a==0&&r%b==0))
            {
                lastcha=chazhi2;
                ansl=l,ansr=r;
            }
            if(chazhi>0)
                l++;
            else
                r++;
            if(l==a&&r==b)
            {
                if(l<32767)
                    l++;
                else
                    r++;
            }
        }
        cout<<ansl<<' '<<ansr<<endl;
    }
}

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