CF - A. Fraction-蒲公英云

CF - A. Fraction

题目连接：http://codeforces.com/contest/854/problem/A

题目描述

Description

Petya is a big fan of mathematics, especially its part related to fractions. Recently he learned that a fraction is called proper iff its numerator is smaller than its denominator (a < b) and that the fraction is called irreducible if its numerator and its denominator are coprime (they do not have positive common divisors except 1).

During his free time, Petya thinks about proper irreducible fractions and converts them to decimals using the calculator. One day he mistakenly pressed addition button ( + ) instead of division button (÷) and got sum of numerator and denominator that was equal to n instead of the expected decimal notation.

Petya wanted to restore the original fraction, but soon he realized that it might not be done uniquely. That’s why he decided to determine maximum possible proper irreducible fraction such that sum of its numerator and denominator equals n. Help Petya deal with this problem.

Input

In the only line of input there is an integer n (3 ≤ n ≤ 1000), the sum of numerator and denominator of the fraction.

Output

Output two space-separated positive integers a and b, numerator and denominator of the maximum possible proper irreducible fraction satisfying the given sum.

Sample Input

Sample Output

1 2

Sample Input

Sample Output

1 3

Sample Input

Sample Output

5 7

解题思路

输入一个数，分成两个数一个做分子a，一个做分母b，a < b，a/b < 1，而且ab互为质数，要求a/b尽可能大。
可以转换成只要ab的最大公因数为1，就互质了。gcd算法求
如果n是偶数的话，分成两个奇数，一定互质，因为两个奇数相近。
偶数的话需要判定。

AC代码

#include<iostream>
using namespace std;
int gcd(int a, int b) {
  return b == 0 ? a : gcd(b, a%b);
}
int main() {
    int n;
    cin >> n;
    if(n%2 == 0) {
        for(int i=n/2-1, j=n/2+1; i>=1; i--, j++){
            if(gcd(i,j) == 1)
            {
                cout << i << " " << j << endl;
                break;
            }
        }
    }else {
        cout << (n-1)/2 << " " << (n+1)/2 << endl;
    }
    return 0;
}