N - Repeating Decimals

N - Repeating Decimals

The decimal expansion of the fraction 1/33 is 0.03, where the 03 is used to indicate that the cycle 03
repeats indefinitely with no intervening digits. In fact, the decimal expansion of every rational number
(fraction) has a repeating 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 cycles are shown below.
Here, we use parentheses to enclose the repeating cycle rather than place a bar over the cycle.
fraction decimal expansion repeating cycle cycle length
1/6 0.1(6) 6 1
5/7 0.(714285) 714285 6
1/250 0.004(0) 0 1
300/31 9.(677419354838709) 677419354838709 15
655/990 0.6(61) 61 2
Write a program that reads numerators and denominators of fractions and determines their repeating
cycles.
For the purposes of this problem, define a repeating cycle of a fraction to be the first minimal length
string of digits to the right of the decimal that repeats indefinitely with no intervening digits. Thus
for example, the repeating cycle of the fraction 1/250 is 0, which begins at position 4 (as opposed to 0
which begins at positions 1 or 2 and as opposed to 00 which begins at positions 1 or 4).
Input
Each line of the input file consists of an integer numerator, which is nonnegative, followed by an integer
denominator, 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 the first occurrence of the cycle
to the right of the decimal or 50 decimal places (whichever comes first), and the length of the entire
repeating cycle.
In writing the decimal expansion, enclose the repeating cycle in parentheses when possible. If the
entire repeating cycle does not occur within the first 50 places, place a left parenthesis where the cycle
begins — it will begin within the first 50 places — and place ‘…)’ after the 50th digit.
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

给出两个数字，计算它们相除的结果，并且算出循环的小数的个数，将循环的小数放在括号之内，当多于50时，用…代替

#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN
int y[3123], c[3111], v[3123];//y记录余数，c记录商，v记录某个余数是第几个出现的
int main()
{
    int n, m;
    while(~scanf("%d %d", &n, &m))
    {
        memset(y, 0, sizeof(y));
        memset(c ,0, sizeof(c));
        memset(v, 0, sizeof(v));
        int top = 1;
        printf("%d/%d = %d", n, m, n/m);
        n = n%m;
        while(n&&v[n]==0)
        {
            v[n] = top;
            y[top] = n;
            n = n * 10;
            c[top++] = n/m;
            n = n % m;
        }
        printf(".");
        for(int i=1;i<top&&i<=50;i++)
        {
            if(n&&n==y[i])
                printf("(");
            printf("%d", c[i]);
        }
        if(n==0)
            printf("(0");
        if(top>50)//余数大于50，用...代替
        {
            printf("...");
        }
        printf(")\n");
        printf("   %d = number of digits in repeating cycle\n\n", n?(top-v[n]):1);//如果a==0,长度为1，否则，长度为余数的总个数-不再循环之列的个数
    }
    return 0;
}