周赛

D 圆周率用acos(-1.0) 使用longlong

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

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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 fromA^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

找规律：每三个数就有两个能被三整除，，找到前A个数中有几个符合要求，再找到前B个有几个符合要求，相减就行：

#include
#include
#include
using namespace std;
int main()
{
	int t,cont=0;
	long long a,b,a1,b1;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld",&a,&b);
		if(a%3==0)
			a1=(a/3)*2-1;
		else
		{
			a1=(a/3)*2+a%3-1;
			if(a%3-1!=0)
				a1=a1-1;
		}
		if(b%3==0)
			b1=(b/3)*2;
		else
		{
			b1=(b/3)*2+b%3-1;
		}
		printf("Case %d: %lld\n",++cont,b1-a1);
	}
	return 0;
}