Codeforces #172 div2的B题

B. Nearest Fraction

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given three positive integers x, y, n. Your task is to find the nearest fraction to fraction  whose denominator is no more than n.

Formally, you should find such pair of integers a, b (1 ≤ b ≤ n; 0 ≤ a) that the value  is as minimal as possible.

If there are multiple "nearest" fractions, choose the one with the minimum denominator. If there are multiple "nearest" fractions with the minimum denominator, choose the one with the minimum numerator.

Input

A single line contains three integers x, y, n (1 ≤ x, y, n ≤ 10⁵).

Output

Print the required fraction in the format "a/b" (without quotes).

Sample test(s)

input

3 7 6

output

2/5

input

7 2 4

output

7/2

题目意思还是很清楚的，要找一个分数使得它与x/y的差最小，要求分母b<n，b要最小，分子也最小的那个。

昨晚纠结了半天，还是没纠结出来，今天把它a了。。。。。。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;double go(double xy,int a,int b){    double s=((double)a)/((double)b)-xy;    if(s>0)return s;    else return -s;}int main(){    int x,y,n;    while(scanf("%d%d%d",&x,&y,&n)!=EOF)    {        double xy=((double)x)/((double)y);        int a,b,j=0;        double ss=110005,v;        for(int i=1;i<=n;i++)        {            while(go(xy,j,i)>go(xy,j+1,i))j++;            v=go(xy,j,i);            if(v<ss)            {                ss=v;                a=j;                b=i;            }            if(ss==0)break;        }        cout<<a<<"/"<<b<<endl;    }    return 0;}