LightOJ 1136 - Division by 3

1136 - Division by 3

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
2 3 5 10 110	Case 1: 2 Case 2: 67

 

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PROBLEM SETTER: JANE ALAM JAN

思路:直接暴力，果断不行，找规律吧，你会发现三个连续的数，只有一个不符合。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	int t,a,b,sum,k;
	scanf("%d",&t);
	k=1;
	while(t--)
	{
		scanf("%d%d",&a,&b);
		int count=(b+1-a)/3;//注意是用b+1 
		int m=(b+1-a)%3;
		sum=count*2;
		if(m==1)
		{
			int x=(b-1)/3;
			if(x*3+1!=b)
			sum++;
		}
		else if(m==2)
		{
			int x=(b-1)/3;
			if(x*3+1!=b)
			sum++;
			int y=(b-1-1)/3;
			if(y*3+1!=b-1)
			sum++;
		 } 
		printf("Case %d: %d\n",k++,sum);
	}
	return 0;
 }