Description
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 fromAth 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.
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 Ath and Bth which are divisible by 3.
Sample Input
2
3 5
10 110
Sample Output
Case 1: 2
Case 2: 67
每三个数中有两个是符合要求的 23 123 12345 123456 12345678 123456789 这是前十个
简化后就是第n个数 只要n能被3整出 那么n和 n-1就是符合的
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
int ans=0;
while(t--)
{
ans++;
long long a,b,cnt=0;
scanf("%lld%lld",&a,&b);
long long a1,b1;
if(a%3==0||a%3==1)
{
a1=(a/3)*2;
}
else
{
if(a%3==2)
{
a1=(a/3)*2+1;
}
}
if(b%3==0||b%3==1)
{
b1=(b/3)*2;
}
else
{
if(b%3==2)
{
b1=(b/3)*2+1;
}
}
if(a%3==0||a%3==2)
{
cnt++;
}
cnt+=b1-a1;
printf("Case %d: %lld
",ans,cnt);
}
return 0;
}