codeforces 281B Nearest Fraction

最新推荐文章于 2018-04-14 12:42:45 发布

ACM2272902662

最新推荐文章于 2018-04-14 12:42:45 发布

阅读量1.5k

点赞数 1

CC 4.0 BY-SA版权

分类专栏： Codeforces

本文链接：https://blog.youkuaiyun.com/yongxingao/article/details/8657565

Codeforces 专栏收录该内容

8 篇文章

订阅专栏

本文探讨了在限定分母范围内的最接近特定分数的整数分数的查找方法，详细介绍了枚举分母并计算分子的过程，以及如何避免数据溢出的问题。通过实例展示了算法的应用，并分析了数据处理中可能出现的挑战。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

You are given three positive integers x, y, n. Your task is to find the nearest fraction to fraction $\text{[math]}$ 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 $\text{[math]}$ 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

  这题目，怎么说呢，思路不多久就想出来了， 枚举分母，求分子就可以，但是却是wa，到最后也没能解决，看数据的时候却发现 中间过程在计算的时候，数据出现了溢出。哎，杯具了。

#include <stdio.h>
#include <string.h>
#include <math.h>
struct num
{
    int zi,mu;
    double val;
}res[1000000];
int cmp(const void *e,const void *f)
{
    struct num *p1=(struct num *)e;
    struct num *p2=(struct num *)f;
    if(fabs(p1->val-p2->val)<=1e-15)
    {
        if(p1->mu>p2->mu)
        {
            return 1;
        }else if(p1->mu<p2->mu)
        {
            return -1;
        }else
        {
            if(p1->zi>p2->zi)
            {
                return 1;
            }else if(p1->zi<p2->zi)
            {
                return -1;
            }else
            {
                return 0;
            }
        }
    }
    if(p1->val>p2->val)
    {
        return 1;
    }else
    {
        return -1;
    }
}
int main()
{
    int x,y,n,s,i,top,t;
    long long int zhong;
    double mod;
    while(scanf("%d %d %d",&x,&y,&n)!=EOF)
    {
        top=0;
        for(i=1;i<=n;i++)
        {
            zhong=((long long int)i*(long long int)x);
            mod=(double)(zhong)/y;
            t=(int)(mod+0.01);
            res[top].mu=i;
            res[top].zi=t;
            res[top++].val=fabs((double)x/(double)y-(double)t/(double)i);
            res[top].mu=i;
            res[top].zi=t+1;
            res[top++].val=fabs((double)x/(double)y-(double)(t+1)/(double)i);
            res[top].mu=i;
            res[top].zi=t+2;
            res[top++].val=fabs((double)x/(double)y-(double)(t+2)/(double)i);
            res[top].mu=i;
            if(t-1>=0)
            {
                res[top].mu=i;
                res[top].zi=t-1;
                res[top++].val=fabs((double)x/(double)y-(double)(t-1)/(double)i);
            }
            if(t-2>=0)
            {
                res[top].mu=i;
                res[top].zi=t-2;
                res[top++].val=fabs((double)x/(double)y-(double)(t-2)/(double)i);
            }
        }
        qsort(res,top,sizeof(res[0]),cmp);
        printf("%d/%d\n",res[0].zi,res[0].mu);
    }
    return 0;
}