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...
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 int main(void) 3 { 4 long n; 5 scanf("%ld",&n); 6 printf("%ld\n",n/3*2+((n%3==2)?1:0)); 7 return 0; 8 }