HDU 2.1.8 小数化分数2

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。

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。

Output 对每一个对应的小数化成最简分数后输出，占一行。

Sample Input 3 0.(4) 0.5 0.32(692307)

Sample Output 4/9 1/2 17/52

Source 2007省赛集训队练习赛（2）

Recommend lcy

#include<stdio.h>  
#include"algorithm"  
using namespace std;  
unsigned long long gcd(unsigned long long a, unsigned long long b)  
{  
    return b == 0 ? a : gcd(b,a%b);  
}  
int main()  
{  
    //freopen("f://in.txt", "r", stdin);  
    int testCase;  
    char inNum[20];  
    scanf("%d", &testCase);  
    getchar();  
    while (testCase--)  
    {  
        memset(inNum, 0, sizeof(inNum));  
        int len = 0, leftBracket = 0, rightBracket=0;  
        scanf("%s", inNum);  
        while (1)  
        {  
            if (inNum[len] == '\0')break;  
            if (inNum[len] == '(')leftBracket = len;  
            if (inNum[len] == ')')rightBracket = len;  
            len++;  
        }  
        if (len == 0)  
            break;  
        unsigned long long numerator = 0, denominator,gcdND;  
        if (leftBracket == 0)//纯小数  
        {  
            for (int i = 0; i < len - 2; i++)  
                numerator += unsigned long long(round((int(inNum[i + 2]) - '0')*pow(10, (len - 3 - i))+0.000001));  
            denominator = unsigned long long(round(pow(10, (len - 2)) + 0.000001));  
        }  
        if (leftBracket == 2)//纯循环小数  
        {  
            int bracketLen = rightBracket - leftBracket - 1;  
            for (int i = 0; i < bracketLen; i++)  
                numerator += unsigned long long(round((int(inNum[i + 3]) - '0')*pow(10, (bracketLen - 1 - i)) + 0.000001));  
            denominator = unsigned long long(round(pow(10, bracketLen) - 1 + 0.000001));  
        }  
        if (leftBracket > 2)//部分循环小数  
        {  
            int bracketLen = rightBracket - leftBracket - 1;  
            int outLen = leftBracket - 2;  
            for (int i = 0; i < outLen; i++)  
                numerator += unsigned long long(round((int(inNum[i + 2]) - '0')*pow(10, (bracketLen + outLen - 1 - i)) + 0.000001));  
            for (int i = 0; i < bracketLen; i++)  
                numerator += unsigned long long(round((int(inNum[i + 3 + outLen]) - 48)*pow(10, (bracketLen - 1 - i)) + 0.000001));  
            for (int i = 0; i < outLen; i++)  
                numerator -= unsigned long long(round((int(inNum[i + 2]) - 48)*pow(10, (outLen - 1 - i)) + 0.000001));  
            denominator = unsigned long long(round(pow(10, (bracketLen + outLen)) - pow(10, outLen) + 0.000001));  
        }  
        gcdND = gcd(denominator, numerator);  
        printf("%lld/%lld\n", numerator / gcdND, denominator / gcdND);  
    }  
    return 0;  
}