1 The decimal expansion of the fraction 1/33 is 0.03, where the 03 is used to indicate that the cycle 03 2 repeats indefinitely with no intervening digits. In fact, the decimal expansion of every rational number 3 (fraction) has a repeating cycle as opposed to decimal expansions of irrational numbers, which have no 4 such repeating cycles. 5 Examples of decimal expansions of rational numbers and their repeating cycles are shown below. 6 Here, we use parentheses to enclose the repeating cycle rather than place a bar over the cycle. 7 fraction decimal expansion repeating cycle cycle length 8 1/6 0.1(6) 6 1 9 5/7 0.(714285) 714285 6 10 1/250 0.004(0) 0 1 11 300/31 9.(677419354838709) 677419354838709 15 12 655/990 0.6(61) 61 2 13 Write a program that reads numerators and denominators of fractions and determines their repeating 14 cycles. 15 For the purposes of this problem, define a repeating cycle of a fraction to be the first minimal length 16 string of digits to the right of the decimal that repeats indefinitely with no intervening digits. Thus 17 for example, the repeating cycle of the fraction 1/250 is 0, which begins at position 4 (as opposed to 0 18 which begins at positions 1 or 2 and as opposed to 00 which begins at positions 1 or 4). 19 Input 20 Each line of the input file consists of an integer numerator, which is nonnegative, followed by an integer 21 denominator, which is positive. None of the input integers exceeds 3000. End-of-file indicates the end 22 of input. 23 Output 24 For each line of input, print the fraction, its decimal expansion through the first occurrence of the cycle 25 to the right of the decimal or 50 decimal places (whichever comes first), and the length of the entire 26 repeating cycle. 27 In writing the decimal expansion, enclose the repeating cycle in parentheses when possible. If the 28 entire repeating cycle does not occur within the first 50 places, place a left parenthesis where the cycle 29 begins — it will begin within the first 50 places — and place ‘...)’ after the 50th digit. 30 Sample Input 31 76 25 32 5 43 33 1 397 34 Sample Output 35 76/25 = 3.04(0) 36 1 = number of digits in repeating cycle 37 5/43 = 0.(116279069767441860465) 38 21 = number of digits in repeating cycle 39 1/397 = 0.(00251889168765743073047858942065491183879093198992...) 40 99 = number of digits in repeating cycle
题目大意：输入，a,b（保证a/b是循环小数），问：从那一段开始循环，非循环部分直接输出，循环部分用括号括起来，但是如果循环部分未在小数点后50位内全部打出，在第50位后输出"...)"
分析：首先我们要先解决拿取小数部分的问题（毕竟如果直接a/b的小数部分会有精度损失），而a%b*10/b（如1/10，1%10*10/10），用这种方法恰好解决了这个问题,接着就是如何判断循环的问题，我个人的想法是看a%b的余数，如果余数相同，该从这位后都以前一部分循环。
PS：注意输出格式，第二行开头要空3格，每个例子间空一行。
#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define f first
#define s second
#define pb push_back
#define dbg(x) cout<<#x<<" = "<<(x)<<endl;
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>PLL;
typedef pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const double inf=1e8+100;
const double eps=1e-8;
const int maxn =1e4+100;
const int N = 1e3+10;
const ll mod=1e9+7;
//------
//define
int arr[maxn];
map<int,int>mp;
//solve
void solve() {
int a,b;
while(cin>>a>>b) {
memset(arr,0,sizeof(arr));
mp.clear();
cout<<a<<"/"<<b<<" = "<<a/b<<".";
a%=b;
int ze=0;
int r,l;
mp[a]=0;
for(int i=1;i<3001;i++){
a*=10;
arr[i]=a/b;
if(mp.count(a%b)){
l=mp[a%b];
r=i;
break;
}else{
mp[a%b]=i;
}
a%=b;
}
int tl=l+1,tr=r;//第l+1才是循环的头
for(int i=1;i<=50&&i<=tr;i++){
if(i==tl)cout<<"(";
cout<<arr[i];
}
if(r<=50)
cout<<")"<<endl<<" "<<r-l;
else{
cout<<"...)"<<endl<<" "<<r-l;
}
cout<<" = number of digits in repeating cycle"<<endl<<endl;
}
}
//main
int main() {
ios_base::sync_with_stdio(false);
#ifdef debug
freopen("in.txt", "r", stdin);
// freopen("out.txt","w",stdout);
#endif
cin.tie(0);
cout.tie(0);
solve();
/*
#ifdef debug
fclose(stdin);
fclose(stdout);
system("out.txt");
#endif
*/
return 0;
}