假期訓練—— Repeating Decimals UVA - 202 模拟

題目連接配接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=138

解題思路：

模拟除法，在每一次進行除法運算時都會有一個被除數，如果兩次被除數相同，則遇到一個循環節，

通過判斷其出現位置，解題。

代碼如下：

#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;	
}