Gym 100499I (数学 模拟)

I. Fraction

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A repeating decimal, also called a recurring decimal, is a number whose decimal representation eventually becomes periodic (i.e., the same sequence of digits repeats indefinitely). The repeating portion of a decimal expansion is conventionally denoted within a pair of brackets so, for example

1 / 6 = 0.16666666... = 0.1(6) = 0.1666(6) = 0.166(66)

Both 0.1(6) or 0.1666(6) or 0.166(66) are correct representation of 1 / 6. Given a recurring decimal representation, your task is to find an irreducible fraction that has that representation.

Input

The first line of input contains the number of tests – T (T ≤ 100). Then T tests follow. Each test is printed in a line as a string whose length does not exceed 15. It is guaranteed to be a meaningful representation of a positive fraction.

Output

For each test, print the result in one line in the format Case #x: a/b

Sample test(s)

input

40.1250.(142857)0.1(6).2

output

Case #1: 1/8Case #2: 1/7Case #3: 1/6Case #4: 1/5

题意是给一个循环小数和循环节，输出对应的分数。

一个循环小数对应的分数可以由两部分相加而来，也就是循环部分加不循环部分。不循环部分很容易求，循环部分我们可以求出它的循环节长度假设是l，设这个循环小数对应的分数是x，设循环节对应的整数部分是y，(比如循环小数0.9898(98)，它的循环节对应的整数部分就是98)，那么就可以列出方程

x*10^l - y = x，通过这个计算就可以求出x，然后对于两部分的分数通分相加，约去gcd就行了。

<span style="font-size:12px;">#include <bits/stdc++.h>using namespace std;#define maxn 111char a[maxn];int n;long long p[22];void solve () {    n = strlen (a);    long long x = 0, y = 0, z = 0;    long long l1 = 0, l2 = 0;    for (int i = 0; i < n; i++) {        if (a[i] == '.') {            i++;            while (i < n && a[i] != '(') {                l1++;                y = y*10 + a[i]-'0';                i++;            }        }        if (a[i] == '(') {            i++;            while (i < n && a[i] != ')') {                l2++;                z = z*10 + a[i]-'0';                i++;            }        }    }    for (int i = 0; i < n && a[i] != '.'; i++) {        x = x*10 + a[i]-'0';    }    if (z == 0) {        long long pp = x*p[l1]+y, qq = p[l1];        long long g = __gcd (pp, qq);        printf ("%lld/%lld\n", pp/g, qq/g);        return ;    }    long long pp = z+(p[l2]-1)*p[l1]*x+(p[l2]-1)*y, qq = p[l1]*(p[l2]-1);    long long g = __gcd (pp, qq);    printf ("%lld/%lld\n", pp/g, qq/g);}int main () {    p[0] = 1;    for (int i = 1; i <= 18; i++)        p[i] = p[i-1]*10;    int t, kase = 0;    scanf ("%d", &t);    while (t--) {        scanf ("%s", a);        printf ("Case #%d: ", ++kase);        solve ();    }    return 0;}</span>