lightoj(1136 - Division by 3 )规律

最新推荐文章于 2019-12-30 15:00:48 发布

原创最新推荐文章于 2019-12-30 15:00:48 发布 · 792 阅读

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本文介绍了一种算法，用于找出特定整数序列中从A到B范围内可以被3整除的整数数量。通过观察序列中每个数字除以3的余数规律，实现了快速计算。

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There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Nowyou are given two integers A and B, you have to find the numberof 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 thesequence are, 123, 1234, 12345. And 123, 12345 are divisible by 3. So, theresult 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 numbersin the sequence between A^th and B^th whichare divisible by 3.

Sample Input

Output for Sample Input

3 5

10 110

Case 1: 2

Case 2: 67

题意

有序列1，12，123，1234，…，12345678910，…现在你是给定两个整数a和b，你必须从第A个找到第B个整数（含），且能被3整除。

例如，让一个= 3。B = 5。因此，序列中的数字是，123，1234，12345。123、12345能被3整除。所以，结果是2。

思路

找规律：1,0,0,1,0,0,1,0,0,1,0,0,1,0,0......

1是不可整除，0是可整除。

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#define in(a,b) scanf("%d%d",&a,&b)
using namespace std;
int main(){
	int i,j,k,m,n,T,ans;
	scanf("%d",&T);k=T;
	while(T--){
		ans=0;
		in(m,n);
		while(m%3!=1){
			m++;
			ans++;
		}
		while(n%3!=1){
			n--;
			ans++;
		}
		ans=ans+(n-m)/3*2;
			printf("Case %d: %d\n",k-T,ans);
}
	return 0;
}