lightoj-1136 - Division by 3【思维】【找规律】

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1136 - Division by 3

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 A^th number to B^th (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.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains two integers A and B (1 ≤ A ≤ B < 2³¹) in a line.

Output

For each case, print the case number and the total numbers in the sequence between A^th and B^th which are divisible by 3.

题解：判断能否被 3 整除，就看各个位上数字之和是否为 3 的倍数。题中第 n 个数各个位上数的和是 n （n + 1）/ 2；所以3 的倍数相邻的数能被 3 整除，然后找规律

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define LL long long
#define M(x) (x*(x+1)/2)
using namespace std;
LL a,b;
int main()
{
  int t,text=0;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%lld %lld",&a,&b);
    LL ans=(b/3-a/3)*2;
    if(a%3==0)  ans++;
    if((b+1)%3==0)  ans++;
    printf("Case %d: %lld\n",++text,ans);
  }
  return 0;
}