SGU 193.Chinese Girls' Amusement

/*

       实际上就是求一个k，满足k<=n/2，且gcd(n,k)=1

       如果n为奇数，k为[n/2]

       如果n为偶数，k=n/2-1-(n/2)%2

*/

#include <iostream>

using namespace std;

string s;

void div2() {

	string t;

	int l = s.size() - 1, tem = s[0] - '0';

	if (tem > 1) t += '0' + tem / 2;

	tem &= 1;

	for (int i = 1; i <= l; i++) {

		tem = tem * 10 + s[i] - '0';

		t += '0' + tem / 2;

		tem &= 1;

	}

	s = t;

}

void cut1() {

	int t = s.size() - 1;

	while (s[t] == '0')

		s[t--] = '9';

	s[t] = s[t] - 1;

}

int main() {

	cin >> s;

	int l = s.size() - 1;

	if ( (s[l] - '0') & 1) {

		cut1(); div2();

		cout << s;

	}

	else {

		div2();

		l = s.size() - 1;

		if ( (s[l] - '0') & 1) cut1();

		cut1();

		cout << s;

	}

	return 0;

}