一、Yet Another Two Integers Problem

本题链接：Yet Another Two Integers Problem

题目：
A. Yet Another Two Integers Problem
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given two integers a and b.

In one move, you can choose some integer k from 1 to 10 and add it to a or subtract it from a. In other words, you choose an integer k∈[1;10] and perform a:=a+k or a:=a−k. You may use different values of k in different moves.

Your task is to find the minimum number of moves required to obtain b from a.

You have to answer t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤2⋅1e4) — the number of test cases. Then t test cases follow.

The only line of the test case contains two integers a and b (1≤a,b≤1e9).

Output
For each test case, print the answer: the minimum number of moves required to obtain b from a.

Example
input
6
5 5
13 42
18 4
1337 420
123456789 1000000000
100500 9000
output
0
3
2
92
87654322
9150

Note
In the first test case of the example, you don’t need to do anything.

In the second test case of the example, the following sequence of moves can be applied: 13→23→32→42 (add 10, add 9, add 10).

In the third test case of the example, the following sequence of moves can be applied: 18→10→4 (subtract 8, subtract 6).

本博客给出本题截图：

题意：给出两个数a和b，每次我们可以把b加上或减去（1~10）中的任何一个数字，问最少操作几步可以把数字b变成数字a

AC代码

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int t;
    cin >> t;
    
    while (t -- )
    {
        int a, b;
        cin >> a >> b;
        
        int t = abs(a - b);
        if (t % 10) 
            cout << t / 10 + 1 << endl;
        else cout << t / 10 << endl;
    }
    
    return 0;
}

  

 

  
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总结

水题，不解释

文章来源: chen-ac.blog.csdn.net，作者：辰chen，版权归原作者所有，如需转载，请联系作者。

原文链接：chen-ac.blog.csdn.net/article/details/117886315