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Problem Description
Ray 在数学课上听老师说，任何小数都能表示成分数的形式，他开始了化了起来，很快他就完成了，但他又想到一个问题，如何把一个循环小数化成分数呢?
请你写一个程序不但可以将普通小数化成最简分数，也可以把循环小数化成最简分数。 |
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Input
第一行是一个整数N，表示有多少组数据。
每组数据只有一个纯小数，也就是整数部分为0。小数的位数不超过9位，循环部分用()括起来。 |
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Output
对每一个对应的小数化成最简分数后输出，占一行。
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Sample Input
3 0.(4) 0.5 0.32(692307) |
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Sample Output
4/9 1/2 17/52 |
View Code
1 #include <stdio.h> 2 #include <math.h> 3 #define MAX 20 4 int gcd(int a, int b)//method:get the greatest divisor number 5 { 6 int r; 7 while (b) 8 { 9 r = a % b; 10 a = b; 11 b = r; 12 } 13 return a; 14 } 15 int main() 16 { 17 int count1, count2, num, j, numerator, denominator, tempa, tempb, temp; 18 bool flag; 19 char c, result[MAX]; 20 scanf("%d", &num); 21 while(num--) 22 { 23 //initialize the parameter 24 flag = false; 25 tempa = 0; 26 tempb = 0; 27 scanf("%s", result); 28 //handle the input, translate the string to double. 29 for (j = 2; result[j] != '\0'; j++) 30 { 31 if (result[j] == '(') 32 { 33 count1 = j; 34 flag = true; 35 } 36 else if (result[j] == ')') 37 { 38 count2 = j; 39 } 40 else if (flag == false) 41 { 42 tempa = 10 * tempa + (result[j] - '0'); 43 tempb = tempa; 44 } 45 else if (flag == true) 46 { 47 tempb = 10 * tempb + (result[j] - '0'); 48 } 49 } 50 //handle the double number respectively 51 if (flag == true) 52 { 53 denominator = pow(10.0, count2 - 3) - pow(10.0, count1 - 2); 54 numerator = tempb - tempa; 55 } 56 else 57 { 58 denominator = pow(10.0, j - 2); 59 numerator = tempa; 60 } 61 //use the method to get the greatest common divisor 62 temp = gcd(denominator, numerator); 63 //print out the result. 64 printf("%d/%d\n", numerator / temp, denominator / temp); 65 } 66 return 0; 67 }
pay attention
firstly, if you want to translate the circulate decimal to fraction. you can use the mothod as follow
A = 0.12(34)
100A = 12.(34)
10000A = 1234.(34)
10000A - 100A = 1222
then you will get the fraction.
secondly, if you can use %s, and gets, you need not use getchar().
thirdly, the complexity of this question is just classify the two aspects clearly.
for the third tip, I finished the one.
you will find a little different, one, add getchar(). two, you must add if (i > 2) continue. three, you need add the '\0'. important!!
View Code
1 #include <stdio.h> 2 #include <math.h> 3 #define MAX 20 4 int gcd(int a, int b)//method:get the greatest divisor number 5 { 6 int r; 7 while (b) 8 { 9 r = a % b; 10 a = b; 11 b = r; 12 } 13 return a; 14 } 15 int main() 16 { 17 int count1, count2, num, i, j, k, numerator, denominator, tempa, tempb, temp; 18 int flag; 19 char c, result[MAX]; 20 scanf("%d", &num); 21 getchar(); 22 k = 0; 23 while(k < num) 24 { 25 //initialize the parameter 26 flag = 0; 27 tempa = 0; 28 tempb = 0; 29 i = 0; 30 while ((c = getchar()) != '\n') 31 { 32 if (c != ' ') 33 { 34 result[i] = c; 35 i++; 36 } 37 } 38 if (i < 2) 39 continue; 40 k++; 41 result[i] = '\0'; 42 //handle the input, translate the string to double. 43 for (j = 2; result[j] != '\0'; j++) 44 { 45 if (result[j] == '(') 46 { 47 count1 = j; 48 flag = 1; 49 } 50 else if (result[j] == ')') 51 { 52 count2 = j; 53 } 54 else if (flag == 0) 55 { 56 tempa = 10 * tempa + (result[j] - '0'); 57 tempb = tempa; 58 } 59 else if (flag == 1) 60 { 61 tempb = 10 * tempb + (result[j] - '0'); 62 } 63 } 64 //handle the double number respectively 65 if (flag == 1) 66 { 67 denominator = pow(10.0, count2 - 3) - pow(10.0, count1 - 2); 68 numerator = tempb - tempa; 69 } 70 else 71 { 72 denominator = pow(10.0, j - 2); 73 numerator = tempa; 74 } 75 //use the method to get the greatest common divisor 76 temp = gcd(denominator, numerator); 77 //print out the result. 78 printf("%d/%d\n", numerator / temp, denominator / temp); 79 } 80 return 0; 81 }