202 - Repeating Decimals

Posted 2021-12-20 guopinghai

#include <stdio.h>
#include <string.h>

using namespace std;

int isExist(int* list, int size, int val);

int main()
    int m, n, divided[5000], count, idx, flag;
    char decimal[5000];
    while(scanf("%d%d", &m, &n) != EOF)
        count = 0, idx = -1;
        memset((void *)divided, ‘\0‘, 500);
        memset(decimal, ‘\0‘, 500);
        printf("%d/%d = %d.", m, n, m / n);
        m = m % n;
        while(true)
            m = m * 10;
            idx = isExist(divided, count, m);
            if(idx != -1)
                break;
            
            divided[count] = m;
            decimal[count] = (m / n) + ‘0‘;
            m = m % n;
            count++;
        
        for(int i = 0; i < idx; ++i)
            if(i >= 50)
                break;
            
            printf("%c", decimal[i]);
        
        printf("(");
        for(int i = idx; i < count; ++i)
            if(i >= 50)
                printf("...");
                break;
            
            printf("%c", decimal[i]);
         
        printf(")\n");
        printf("   %d = number of digits in repeating cycle\n\n", count - idx);
    
    



int isExist(int* list, int size, int val)
    for(int i = 0; i < size; ++i)
        if(list[i] == val)
            return i;
        
    
    return -1;


// 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

//https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=830&page=show_problem&problem=138

//还是要注意整除与取余在数字运算中的重要作用
//代码能力！！！