Div 3

Div3 

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1136 uDebug

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 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.

Sample Input

2

3 5

10 110

Sample Output

Case 1: 2

Case 2: 67

思路：打表直接会超时，换种方法做；

打出前十几个数会发现这样的一个规律：0 1 2 2 3 4 4 5 6 6 7 8 8……，这时就需要找规律来优化算法；

可以以3为一个周期，则从0到a[i]被3整除的个数为若a[i]被3整除是a[i]/3*2，否则为a[i]/3*2+a[i]%3-1;

代码：

#include<stdio.h>int f(int x){int a=x/3;if(x%3==0)return a*2;elsereturn a*2+x%3-1; }int main(){int t,a,b,mm=1;scanf("%d",&t);while(t--){scanf("%d%d",&a,&b);int c=f(b)-f(a-1);printf("Case %d: %d\n",mm++,c);}return 0; }