【LightOJ】1136 - Division by 3（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 Ath number toBth (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.

被三整除有个规律是和可以被三整除，那么我们观察一下数字和是怎么变化的。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
#define LL long long
#define PI acos(-1.0)
int cal(int x)
{
	int a = x / 3;
	int b = x % 3;
	if (b != 0)
		return a * 2 + b - 1;
	else
		return a * 2;
}
int main()
{
	int u;
	int Case = 1;
	int l,r;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d %d",&l,&r);
		printf ("Case %d: ",Case++);
		int ans = 0;
		ans += cal(r) - cal(l-1);
		printf ("%d\n",ans);
	}
	return 0;
}