1136 - Division by 3

题意:There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Now you are given two integers A and B, you have to find the number of integers from Ath number to Bth (inclusive) number, which are divisible by 3.
For example, let A = 3. B = 5. So, the numbers in the sequence are, 123, 1234, 12345. And 123, 12345 are divisible by 3. So, the result is 2.

思路：找到规律，1 0 0 1 0 0 1 0 0......（1不能被3整除，0能被3整除），然后就是细节问题，但还是w了好几次.......- -！

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1136

 

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;

int main(){
    
//    freopen("data.in","r",stdin);
//    freopen("data.out","w",stdout);
    
    int t,a,b,i;
    scanf("%d",&t);
    for(i=1;i<=t;i++){
        scanf("%d%d",&a,&b);
        int sum=0;
        if(a%3!=1){
            if(a%3==2){sum+=2;a+=2;}
            else {sum++;a++;}    
        }
        if(b%3!=1){
            if(b%3==2) {sum++;b--;}
            else {sum+=2;b-=2;}
        }
        printf("Case %d: %d\n",i,sum+b-a-(b-a)/3);
    }
    return 0;
}