A - Add Odd or Subtract Even

You are given two positive integers aa and bb.In one move, you can change aa in the following way:

• Choose any positive odd integer x(x>0) and replace a with a+x;

• choose any positive even integer y (y>0) and replace aa with a−y.

You can perform as many such operations as you want. You can choose the same numbers x and y in different moves.

Your task is to find the minimum number of moves required to obtain b from a. It is guaranteed that you can always obtain b from a.

You have to answer tt independent test cases.

Input
The first line of the input contains one integer t (1≤t≤104) — the number of test cases.
Then tt test cases follow. Each test case is given as two space-separated integers a and b (1≤a,b≤109).

Output
For each test case, print the answer — the minimum number of moves required to obtain bbfrom aa if you can perform any number of moves described in the problem statement. It is guaranteed that you can always obtain b from a.

ExampleInput
5
2 3
10 10
2 4
7 4
9 3

Output
1
0
2
2
1

题意
给你两个数a,b,通过以下方式更改a：
选择任意正奇数x (x>0)，将a替换为a+x；
选择任意正偶数y (y>0)，将a替换为a-y。
求从a中得到b所需的最少移动次数

思路
判断a,b大小，求差，判奇偶

代码

#include
using namespace std;
int a,b,c,n;
int main()
{
 scanf("%d",&n);
 while(n--)
 {
  scanf("%d%d",&a,&b);
  if(b==a)
   printf("0\n");
  else if(b>a)
  {
   c=b-a;
   if(c%2)//c是奇数，处理一次，加上一个奇数
    printf("1\n");
   else //c是偶数，处理两次，加上两个奇数
    printf("2\n");
  }
  else if(b<a)
  {
   c=a-b;
   if(c%2==0)//c是偶数，处理一次，减去一个偶数
    printf("1\n");
   else //c是奇数，处理两次，减去一个偶数，加上一个奇数
    printf("2\n");
  }
 }
 return 0;
}