题目连接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=138
解题思路：
模拟除法，在每一次进行除法运算时都会有一个被除数，如果两次被除数相同，则遇到一个循环节，
通过判断其出现位置，解题。
代码如下：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(int i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
typedef vector<int> vi;
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
const int N = 1e7+5;
int a,b;
string re;
int head1;
int c[N];
int index1[N];
int aa,bb;
int main()
{
while(cin >> a>>b)
{
aa = a;
bb = b;
memset(c,0,sizeof(c));
memset(index1,0,sizeof(index1));
re.clear();
head1 = a/b;
a %= b;
int ind1,ind2;
while(1)
{
if(c[a])
{
ind1 = index1[a];
ind2 = re.size()-1;
break;
}
else
{
c[a]++;
index1[a] = re.size()-1;
}
a*=10;
re += a/b+'0';
a %= b;
}
//cout << re<<endl;
int ctt = 0;
int fff = 0;
cout << aa<<"/"<<bb<<" = "<<head1<<".";
for(int i=0;i<=ind1;i++)
{
cout << re[i];
}
cout << "(";
for(int i=ind1+1;i<=ind2;i++)
{
ctt++;
cout << re[i];
if(ctt>=50)
{
fff = 1;
break;
}
}
if(fff)
{
cout <<"...";
}
cout <<")"<<endl;
cout <<" "<< ind2-ind1<<" = number of digits in repeating cycle"<<endl<<endl;
}
return 0;
}