SGU105 - Div 3

105. Div 3

time limit per test: 0.5 sec. 
memory limit per test: 4096 KB

 

There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3.

 

Input

Input contains N (1<=N<=2³¹ - 1).

 

Output

Write answer to the output.

 

Sample Input

4

Sample Output

2
题解：规律题。。。。可以发现前N项的余数为1,0,0,1,0,0,1,0,0...

#include<stdio.h>
int main(void)
{
    long n;
    scanf("%ld",&n);
    printf("%ld\n",n/3*2+((n%3==2)?1:0));
    return 0;
}