Problem 26 Reciprocal cycles （模拟分数除法）

最新推荐文章于 2022-06-29 20:22:21 发布

原创最新推荐文章于 2022-06-29 20:22:21 发布 · 573 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Problem 26 #Reciprocal cycles #模拟分数除法

Project Euler 同时被 3 个专栏收录

81 篇文章

订阅专栏

ACM_STL题

31 篇文章

订阅专栏

ACM_模拟

16 篇文章

订阅专栏

Reciprocal cycles

Problem 26

A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

¹/₂ = 0.5
¹/₃ = 0.(3)
¹/₄ = 0.25
¹/₅ = 0.2
¹/₆ = 0.1(6)
¹/₇ = 0.(142857)
¹/₈ = 0.125
¹/₉ = 0.(1)
¹/₁₀ = 0.1

Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that ¹/₇ has a 6-digit recurring cycle.

Find the value of d < 1000 for which ¹/_d contains the longest recurring cycle in its decimal fraction part.

题解：模拟除法。

代码：

#include <iostream>
#include <set>
using namespace std;
int main()
{
    set<int> s;
    int num,k,flag;
	int ans = 0 ,ANS = 0 , maxn ;
    num = 1 ;
    for (int i = 1 ; i < 1000 ; i++ )
    {
        num = 1; //分子 
        k = i;  //分母 
        flag = 1;
        s.clear();
        
        //模拟 
        while (flag)
        {
        	if ( num < k )
            num *= 10 ;
        //find():返回给定值的定位器，如果没找到则返回end()。
        	flag = s.find(num%k) == s.end() ? 1: 0;
        //没找到就 insert 
        	if (flag) s.insert(num%k);
        	num = num%k ;
        }
        //统计个数 
        ans = s.size();
        if (ans > ANS )
        {
            ANS = ans ;
            maxn = i ;
        }
    }
    cout<<maxn<<endl;
    return 0;
}