Yet Another Two Integers Problem

文章目录

一、Yet Another Two Integers Problem

总结

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;
}

总结

水题，不解释