POJ 1650 Integer Approximation （追赶法求解）

Integer Approximation

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5458 Accepted: 1806

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

3.1415926535897910000

Sample Output

355 113

题意：在L的范围内寻找两个数N ,D，使得fabs(A-N/D)最小

思路：看到别人用的追赶法，看了一下解释，写了一下，复杂度O（n）

总结：也想过二分枚举，但是T了

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 60100#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define mod 1000000007using namespace std;double a;double dis(int n,int d){return a-(double)n/d;}int main(){while(scanf("%lf",&a)!=EOF){int l;scanf("%d",&l);int d=1,n=1;int ansd=d,ansn=n;double mi=INF*1.0;while(d<=l&&n<=l){double k=dis(n,d);//printf("%lf\n",k);double kk=fabs(k);if(kk<mi){mi=kk;ansd=d;ansn=n;}if(k>0)n++;elsed++;}printf("%d %d\n",ansn,ansd);}return 0;}