UVA 202 循环点
| Repeating Decimals |
The decimal expansion of the fraction 1/33 is , wherethe is used to indicate that the cycle 03 repeatsindefinitely with no intervening digits. In fact, the decimal expansion ofevery rational number (fraction) has arepeating cycle as opposed to decimal expansions of irrational numbers,which have no such repeating cycles.
Examples of decimal expansions of rational numbers and their repeating cyclesare shown below. Here, we useparentheses to enclose the repeating cycle rather than place a bar overthe cycle.
Write a program that reads numerators and denominators of fractions anddetermines their repeating cycles.
Forthe purposes of this problem, define a repeating cycle of a fraction to bethe first minimal length string of digits tothe right of the decimal that repeats indefinitely with no intervening digits.Thus for example, the repeating cycle ofthe fraction 1/250 is 0, which begins at position 4 (as opposed to 0 whichbegins at positions 1 or 2 and as opposedto 00 which begins at positions 1 or 4).
Input
Each line of the input file consists of an integer numerator, which isnonnegative, followed by an integerdenominator, which is positive. None of the input integers exceeds 3000.End-of-file indicates the end of input.
Output
For each line of input, print the fraction, its decimal expansion through thefirst occurrence of the cycle to the rightof the decimal or 50 decimal places (whichever comes first), and the lengthof the entire repeating cycle.
In writingthe decimal expansion, enclose the repeating cycle in parentheses when possible.If the entire repeating cycle doesnot occur within the first 50 places, place a left parenthesis where thecycle begins - it will begin within the first 50places - and place ``...)" after the 50th digit.
Print a blank line after every test case.
Sample Input
76 25
5 43
1 397
Sample Output
76/25 = 3.04(0)
1 = number of digits in repeating cycle
5/43 = 0.(116279069767441860465)
21 = number of digits in repeating cycle
1/397 = 0.(00251889168765743073047858942065491183879093198992...)
99 = number of digits in repeating cycle
//第一，查找循环点，如果被除数是m的话，那么他的余数只能是0~m-1。所以通过查找商和余数就可以找到循环点
//拿5/43举个例子
//商： 0 1 1 6 2 7 9 0 6 9 7 6 7 4 4 1 8 6 0 4 6 5 1
//余数： 5 7 27 12 34 39 3 30 42 33 29 32 19 18 8 37 26 2 20 28 22 5 7
//其中第一组为小数点前边的值，最后一组跟前边是重合，那么就表明循环点出现了
#include
#include
#include
using namespace std;
#define maxn 3005
int r[maxn]; //余数
int s[maxn]; //商
int judge[maxn];
int main()
{
int a,b;
while(scanf("%d %d",&a,&b))
{
int m=a;
int n=b;
int count=0;
memset(r,0,sizeof(r));
memset(s,0,sizeof(s));
memset(judge,0,sizeof(judge));
s[count++]=a/b;
a=a%b;
while(a!=0 && judge[a]==0) //n不为0 或者没有找到循环节点
{
judge[a]=1;
s[count] = a*10/b;
r[count] = a*10%b;
//printf("%d %d\n",s[count],r[count]);
count++;
a=a*10%b;
//printf("a=%d,judge[a]=%d\n",a,judge[a]);
}
// printf("%d\n",count);
printf("%d/%d = ",m,n);
if(a == 0)
{
printf("%d.",s[0]);
for(int i=1;i