[BestCoder Round #5] hdu 4956 Poor Hanamichi (数学题）

Posted 2020-09-09 blfbuaa

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:

The odd dights are  and  are even digits.

Hanamichi comes up with a solution, He notices that:
 mod 11 = -1 (or 10),  mod 11 = 1, 
So X mod 11 
= 
= 
= 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 ≤ )

 

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

3
3 4
2 50
7 83

 

Sample Output

-1
-1
80

解题思路：

题意为：存在一个问题求一个区间（[l,r])内有多少数满足（偶数位的和-奇数位的和==3）。

Hanamichi提出了它的解法 Answer =  (r + 8) / 11 – (l – 1 + 8) / 11.我们的任务就是找到一个最小的右端点，使得该区间内满足条件的数不满足Hanamichi的解法。

一開始读完题意想的是数位Dp，但后来用暴力搞了.....

代码：

#include <iostream>
using namespace std;

long long cal(long long l,long long r)//常规方法求区间内满足条件的数有多少个
{
    long long result=0;
    for(;l<=r;l++)
    {
        long long temp=l;
        int mul=1;
        int re=0;
        while(temp)
        {
            re+=mul*(temp%10);
            temp/=10;
            mul=(-mul);
        }
        if(re==3)
            result++;
    }
    return result;
}

long long re(long long l,long long r)
{
    return (r+8)/11-(l-1+8)/11;
}

int main()
{
    long long l,r,R;
    int t;
    cin>>t;
    while(t--)
    {
        cin>>l>>r;
        for(R=l;R<=r;R++)//从小到大，输出最小的那个
        {
            long long result=re(l,R);
            if(cal(l,R)!=result)
            {
                cout<<R<<endl;
                break;
            }
        }
        if(R>r)
            cout<<-1<<endl;
    }
    return 0;
}