best code-1001Poor Hanamichi

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Hanamichi参加编程比赛，面对特殊题目：找出指定区间内奇数位之和小于偶数位之和且差值为3的整数。通过模11的方法尝试解答，但存在缺陷。Rukaw邀请你修改测试数据，让Hanamichi的解决方案失效。

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Poor Hanamichi

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 743 Accepted Submission(s): 275

Problem Description

  
   Hanamichi is taking part in a programming contest, and he is assigned to solve a special problem as follow: Given a range [l, r] (including l and r), find out how many numbers in this range have the property: the sum of its odd digits is smaller than the sum of its even digits and the difference is 3.

A integer X can be represented in decimal as:

   
   
    
     X=A n ×10 n +A n−1 ×10 n−1 +…+A 2 ×10 2 +A 1 ×10 1 +A 0  
    


The odd dights are 
   
   
    
     A 1 ,A 3 ,A 5 … 
    

 and 
   
   
    
     A 0 ,A 2 ,A 4 … 
    

 are even digits.

Hanamichi comes up with a solution, He notices that:

   
   
    
     10 2k+1  
    

 mod 11 = -1 (or 10), 
   
   
    
     10 2k  
    

 mod 11 = 1, 
So X mod 11 
= 
   
   
    
     (A n ×10 n +A n−1 ×10 n−1 +…+A 2 ×10 2 +A 1 ×10 1 +A 0 )mod11 
    


= 
   
   
    
     A n ×(−1) n +A n−1 ×(−1) n−1 +…+A 2 −A 1 +A 0  
    


= sum_of_even_digits – sum_of_odd_digits
So he claimed that the answer is the number of numbers X in the range which satisfy the function: X mod 11 = 3. He calculate the answer in this way : 
Answer =  (r + 8) / 11 – (l – 1 + 8) / 11.

Rukaw heard of Hanamichi’s solution from you and he proved there is something wrong with Hanamichi’s solution. So he decided to change the test data so that Hanamichi’s solution can not pass any single test. And he asks you to do that for him.

Input

  
   You are given a integer T (1 ≤ T ≤ 100), which tells how many single tests the final test data has. And for the following T lines, each line contains two integers l and r, which are the original test data. (1 ≤ l ≤ r ≤ 
   
   
    
     10 18  
    

)

Output

  
   You are only allowed to change the value of r to a integer R which is not greater than the original r (and R ≥ l should be satisfied) and make Hanamichi’s solution fails this test data. If you can do that, output a single number each line, which is the smallest R you find. If not, just output -1 instead.

Sample Input

Sample Output

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#define mem(x)  memset((x),0,sizeof((x)))
using namespace std;

int a[20];
int len = 0;
int main()
{
    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);

    int T;
    cin>>T;
    long long l,r;
    while(T--)
    {
        //scanf("%lld%lld",&l,&r);
        cin>>l>>r;
        long long ans = -1;
        for(long long i = l; i <= r; i++)
        {
            long long temp = i;
            if(temp % 11 == 3)
            {
                int k = 0;
                mem(a);
                while(temp)
                {
                    a[k++] = temp%10;
                    temp /= 10;
                }

                int odd = 0; //ÆæÊý
                int even = 0;
                for(int j = 0; j < k; j += 2)
                {
                    odd += a[j];
                    even += a[j+1];
                }

                if(odd - even!= 3)
                {
                    ans = i;
                    break;
                }
            }
        }
        cout<< ans <<endl;
    }
    return 0;
}

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