快速切题 sgu105. Div 3 数学归纳 数位+整除 难度:0

105. Div 3

time limit per test: 0.25 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

实际用时:4h58min
应用时:3min
思路:小学奥赛应该有被3整除需各位之和为3倍数这一点,又被9余数同数位和的9余数,所以可证被3余同数位和的3余数,然后就是,余数遵循1,0,0,1,0,0,....

#include <cstdio>
#include <cstring>
using namespace std;
long long n;
int main(){
    while(scanf("%I64d",&n)==1){
        long long l=(n+2)/3;
        printf("%I64d\n",n-l);
    }
    return 0;
}