本文迁移自老博客，原始链接为 https://seven.blog.ustc.edu.cn/usacofractions-to-decimals/
将分数转换为小数，如果是有限小数，则直接输出结果，若果是循环小数，则找到循环节并用括号括起来表示后面的循环部分。规则如下： 1/3 = 0.(3) 22/5 = 4.4 1/7 = 0.(142857) 2/2 = 1.0 3/8 = 0.375 45/56 = 0.803(571428) 解法：模拟除法，找到循环节的关键是出现重复的余数。
#include<iostream>
#include<algorithm>
#include<fstream>
#include<map>
#include<stdio.h>
#include<string.h>
using namespace std;
ofstream fout ("fracdec.out");
ifstream fin ("fracdec.in");
int n,d ,z ,r;
map<int , int> bt;
int ans[1000000];
int s = 1;
bool flag = false;
int pos;
void div()
{
bt[r] = s;
while(r / d == 0)
{
r *= 10;
if(bt[r])
{
flag = true;
ans[s++] = 0;
pos = bt[r];
return;
}
ans[s++] = 0;
}
ans[s-1] = r / d;
r = r % d;
if(r == 0)
return;
if(bt[r])
{
flag = true;
pos = bt[r];
return;
}
else
div();
}
void output()
{
fout<<z<<".";
int count = 0;
if(z == 0)
count = 1;
while(z)
{
count++;
z /= 10;
}
count++;
if(flag)
{
for(int i = 1 ; i < s; i++)
{
if(i == pos)
{
fout<<"(";
count++;
if(count > 75)
{
fout<<endl;
count = 0;
}
}
fout<<ans[i];
count++;
if(count > 75)
{
fout<<endl;
count = 0;
}
}
fout<<")"<<endl;
}
else{
for(int i = 1 ; i < s; i++)
fout<<ans[i];
fout<<endl;}
}
int main()
{
fin>>n>>d;
z = n / d;
r = n % d;
if(r != 0)
{
div();
output();
}
else
fout<<z<<"."<<0<<endl;
}
