ACM学习历程—HDU1717 小数化分数2（gcd）

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <string>
#define LL long long

using namespace std;

LL e[11], val1, val2, A, B;
int len1, len2;
char str[50];

void Init()
{
    e[0] = 1;
    for (int i = 1; i < 11; ++i)
    {
        e[i] = 10*e[i-1];
    }
}

void Input()
{
    scanf("%s", str);
    len1 = len2 = 0;
    val1 = val2 = 0;
    for (int i = 2; str[i] != ‘\0‘; ++i)
    {
        if (str[i] == ‘(‘)
        {
            for (int j = i+1; str[j] != ‘)‘; ++j)
            {
                val2 = 10*val2 + str[j] - ‘0‘;
                len2++;
            }
            break;
        }
        val1 = 10*val1 + str[i] - ‘0‘;
        len1++;
    }
    if (len2)
    {
        A = val1*(e[len2] - 1) + val2;
        B = e[len1] * (e[len2]-1);
    }
    else
    {
        A = val1;
        B = e[len1];
    }
}

LL gcd(LL a, LL b)
{
    if (b == 0)
        return a;
    else
        return gcd(b, a%b);
}

void Work()
{
    LL d = gcd(A, B);
    A /= d;
    B /= d;
    printf("%I64d/%I64d\n", A, B);
}

int main()
{
    //freopen("test.in", "r", stdin);
    int T;
    scanf("%d", &T);
    Init();
    for (int times = 0; times < T; ++times)
    {
        Input();
        Work();
    }
    return 0;
}