Sgu/105

找规律即可

#include<iostream>
 
using namespace std;
 
int main()
{
	int n;
	cin >> n;
	cout << 2 * (n / 3) + ( n % 3 > 0 ? ( (n % 3)- 1 ) : 0) << endl;
	return 0;
}

同，规律

Program Div_3;
Var
	n:longint;
Begin
	readln(n);
	writeln(trunc(n/3*2));
End.

#include <cstdio>
using namespace std;
 
int main()
{
	int N;
	scanf("%d", &N);
	printf("%d\n", (N / 3 << 1) + (N % 3 == 2));
 
	return 0;
}