这是群赛上的一道题目，是道数论题。题意比较简单，就是给你一种这样的数，1，12，123，1234，12345，123456.。。。。。。求第a个这样的数到第b个这样的数之间，有多少个数能被3整除。思路很简单，第n个数能否被3整除，只需要从1一直加到n即可。注意，这里的加，是指把某个数的每一位加起来，这是根据一个数能否被3整除的性质所决定的。但是若要循环的话，由于数据范围太大，会超时。
我刚开始把前100项能被3整除的个数打表打出来后，发现规律很明显，之后就1A了。规律是这样的，能被3整除的个数是这样分布的，0，1，2，2，3，4，4，5，6，6，7，8，8.。。。。。。即奇数出现一次，偶数出现两次，，我们可以以3为周期，接下来就很简单了。题目：
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 Ath number to Bth (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.
Sample Input
2
3 5
10 110
Sample Output
Case 1: 2
Case 2: 67
ac代码：
#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
int fun(int x)
{
int a = x/3;
int b = 2*a;
if(x%3 == 2)
{
b+=1;
}
return b;
}
int main()
{
int numcase;
scanf("%d",&numcase);
for(int k = 1;k <= numcase;++k)
{
int m,n,s = 0,e = 0;
scanf("%d%d",&m,&n);
if(m == 1 || m == 2) s = 0;
if(n == 1) e = 0;
else if(n == 2) e = 1;
else
{
s = fun(m-1);
e = fun(n);
}
printf("Case %d: ",k);
printf("%d\n",e-s);
}
return 0;
}