0/7 =13
2
#include <iostream> #include <stdio.h> #include <stdlib.h> using namespace std; bool existRemainder(int source[], int n); int main() { int Calculate(const int numerator, const int denominator, char *decimaltostring); int n, m; char p[103]; cin >> n >> m; cout << Calculate(n, m, p) << endl; cout << p << endl; return 0; } int Calculate(const int numerator, const int denominator, char *decimaltostring) { if (denominator == 0) return -1; if (numerator % denominator == 0) { itoa(numerator / denominator, &(decimaltostring[0]), 10); //整数转为字符串 return 0; } int remainder[110] = { 0 }; //保存余数 int decimal[110] = { 0 }; //保存小数部分 remainder[0] = numerator % denominator; itoa(numerator / denominator, &(decimaltostring[0]), 10); decimaltostring[1] = '.'; int rlable = 0; int dlable = 0; int i, j; while (1) { if (dlable == 100) { remainder[rlable] = -1; //置为-1,未整除但小数超过100位，也不加括号，题目中没说清楚 break; } decimal[dlable++] = remainder[rlable] * 10 / denominator; remainder[rlable + 1] = remainder[rlable] * 10 % denominator; rlable++; if (remainder[rlable - 1] * 10 % denominator == 0)// if(!remainder[rlable])即可 { remainder[rlable] = -1; //置为-1标志整除 break; } if (existRemainder(remainder, rlable)) break; } if (remainder[rlable] == -1) { for (i = 0, j = 2; i < dlable; i++, j++) { itoa(decimal[i], &(decimaltostring[j]), 10); } } else { decimaltostring[2] = '('; for (i = 0, j = 3; i < dlable; i++, j++) { itoa(decimal[i], &(decimaltostring[j]), 10); } decimaltostring[j] = ')'; } return 0; } bool existRemainder(int source[], int n) { int i; for (i = 0; i < n; i++) { if (source[n] == source[i]) return true; } return false; } //实现的想法：保存每次的余数，然后用 余数*10 再除以除数，得到余数后，在所保存的余数数组中查找，如果没有，则继续，如果已经出现了该余数，说明接下来会出现循环了，则跳出循环。文章来源: 华为oj--分数转小数