A. Yet Another Two Integers Problem
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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]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⋅104，1≤t≤2⋅104) — 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≤109，1≤a,b≤109).
Output
For each test case, print the answer: the minimum number of moves required to obtain b from a.
Example
input
Copy
6
5 5
13 42
18 4
1337 420
123456789 1000000000
100500 9000
output
Copy
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).
题意：用多组k把a变成b，问需要最少的操作数，要求最少的操作数，由贪心可得：那么就是直接由最大的k值去求解即可。剩余部分即是其它的k值去进行单个补位而已。
比较复杂的代码：
#include<bits/stdc++.h>
using namespace std;
long long a[15];
int main()
{
cin.tie(0);std::ios::sync_with_stdio(false);//清除cin的缓存时间，减少输入所消耗的时间,就算没有也不会影响AC的
int t;
cin>>t;
while(t--)
{
long long a,b;
cin>>a>>b;
long long flag=abs(a-b);
for(long long i=10;i>=1;i--)
{
long long k=flag/i;
a[i]=k;
flag-=k*i;
}
long long sum=0;
for(long long i=1;i<=10;i++)
{
if(a[i])
sum+=a[i];
}
cout<<sum<<endl;
}
return 0;
}
其实通过发现规律就可以知道，a和b的变换用的基本是10，和其它数字，故而可以简化上述代码。
所以有了下面的简化代码：
#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while (t--)
{
int a, b;
cin>>a>>b;
int flag = abs(a - b);
if (flag % 10 == 0)
{
printf("%d\n", flag / 10);//10的个数
}
else
{
printf("%d\n", flag / 10 + 1);//+1为其它位的数字
}
}
return 0;
}