UVa 202 Repeating Decimals

最新推荐文章于 2020-02-04 20:59:46 发布

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最新推荐文章于 2020-02-04 20:59:46 发布

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分类专栏： ACM 文章标签： uva

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本文详细介绍了如何从输入的整数分子和正整数分母中，计算并输出对应的分数及其十进制展开形式，包括循环节的长度。通过实例演示了完整的算法流程，确保结果准确无误。

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Input

Each line of the input le 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- le indicates the end

of input.

Output

For each line of input, print the fraction, its decimal expansion through the rst occurrence of the cycle

to the right of the decimal or 50 decimal places (whichever comes rst), 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 rst 50 places, place a left parenthesis where the cycle

begins | it will begin within the rst 50 places | and place `

...)

' after the 50th digit.

SampleInput

76 25

5 43

1 397

SampleOutput

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

#include <iostream>
#include <cstdlib>
#include <string.h>
#include <cstdio>

int s[3001];
int m[3001];

int main()
{
    int a, b;
    int temp;
    while(scanf("%d %d", &a, &b) == 2)
    {
    	temp = a;
    	int flag = 1;
    	int count = 0;
    	int num = 1;
    	s[0] = a / b;  //整数数列 
    	m[0] = a % b;  //余数数列 
    	printf("%d/%d = ", a, b);
    	printf("%d.", s[0]);
    	if(m[0] != 0)
    	{	
			for(int i = 0; flag == 1; i++)
			{
				a = m[i];
				a = a * 10;
				s[i+1] = a / b;
				m[i+1] = a % b;
				for(int j = 0; j < i+1; j++)
				{
					if(m[i+1] == m[j])
					{
						count = j+1;
						flag = 0;
						break;
					}
				}
				num++;
			}
			for(int i = 0,j = 0; i < num-1; i++, j++)
			{
				if(j > 49 && i < num - 1)
				{
					printf("...");
					break;
				}
				if(i == count-1)
				  printf("(");
				printf("%d", s[i+1]);
			}
			printf(")\n");
			printf("   %d = number of digits in repeating cycle\n\n", num - count);
		}
		else
		{
			printf("(%d)\n", 0);
			printf("   %d = number of digits in repeating cycle\n\n", 1);
		}		
	}	
	return 0;	
}