[North Central NA Contest 2018] Rational Ratio | 规律 细节模拟

Description

Every (positive) rational number can be expressed as a ratio of two (positive) integers. However, in decimal form, rational numbers often have an infifinitely repeating pattern, e.g., 1/7 = 0.142857142857142857… A convenient way of writing this repeating pattern is to put a bar over the fifirst occurrence of the repeating part, so 1/7 would be written:

Given a rational number consisting of a series of digits, followed by a decimal point, followed by more digits, and then a number indicating how many of the rightmost digits repeat (i.e., the number of digits under the bar), your task is to find the ratio of two integers, in the most reduced form, that represent the same rational number. For example, for the input “0.142857 6” you should fifind 1/7.

Input

The input will be a single line with two numbers separated by one space. The fifirst number will consist of 1 to 3 digits (0–9), followed by a decimal point, followed by 1 to 11 digits (0–9), representing the decimal form of the number, possibly with leading zeros. The second number will be a positive integer indicating how many of the rightmost digits of the preceding number repeat. The fifirst number will always be greater than 0. The second number will never be less than 1 nor larger than the number of digits to the right of the decimal point.

Output

Print the corresponding fraction in its most reduced form, that is, the fraction with the smallest possible integer values in the numerator and denominator.

Samples

Input Copy

0.142857 6

Output

1/7

Input Copy

1.6 1

Output

5/3

Input Copy

123.456 2

Output

61111/495

题意：

将一个无线循环浮点数 转换为分数的形式，给定循环的位数

规律：

[该规律将附到做题总结第26条] 传送门

先只是考虑小数点后的部分：

对于一个浮点数 0.abcdef，假如它的循环节长度为6，即循环节为abcdef，那么可以将其转换为 abcdef/999999，循环节的长度是多少，就除以一个长度为多少的每位为9的数

假如小数点后面的数字不全属于循环节的部分，那么说我们就可以将不属于循环节的部分挪到整数部分，然后最终再把结果除一下即可

string s, t;
int n, flag;
ll zheng, pre, jie;
ll divid;
int siz;
int main() {
  cin >> s >> n;
  int len = s.size();
  for (int i = 0; i < len; i++) {
    if (s[i] == '.') {
      flag = 1;
      continue;
    }
    if (!flag) {
      zheng *= 10;
      zheng += s[i] - '0';
    }
    else {
      t.push_back(s[i]);
    }
  }
  int sizFen = t.size();
  for (int i = sizFen - n; i < sizFen; i++) {
    jie *= 10;
    jie += t[i] - '0';
  }
  int lenPre = sizFen - n;
  for (int i = 0; i < sizFen - n; i++) {
    pre *= 10;
    pre += t[i] - '0';
  }
  for (int i = 1; i <= lenPre; i++) {
    zheng *= 10;
  }
  zheng += pre;
  ll fenzi = 0, fenmu = 0;
  fenzi = jie;
  for (int i = 1; i <= n; i++) {
    fenmu *= 10;
    fenmu += 9;
  }
  fenzi += zheng * fenmu;
  //right pos:
  for (int i = 1; i <= lenPre; i++) {
    fenmu *= 10;
  }
  ll gd = gcd(fenzi, fenmu);
  cout << fenzi / gd << '/' << fenmu / gd << endl;
  return 0;
}
/**
**/