Codeforces Round #624 (Div.3)A. Add Odd or Subtract Even

A. Add Odd or Subtract Even
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given two positive integers a and b.

In one move, you can change a 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 a 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 t independent test cases.

Input
The first line of the input contains one integer t (1≤t≤104) — the number of test cases.

Then t 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 b from a if you can perform any number of moves described in the problem statement. It is guaranteed that you can always obtain b from a.

Example
inputCopy
5
2 3
10 10
2 4
7 4
9 3
outputCopy
1
0
2
2
1
Note
In the first test case, you can just add 1.

In the second test case, you don’t need to do anything.

In the third test case, you can add 1 two times.

In the fourth test case, you can subtract 4 and add 1.

In the fifth test case, you can just subtract 6.

题意：就是给了a和b两个数，现在有两个操作
一是选择奇数 让a变成a+x
二是选择偶数 让a变成a-y
问你最少操作多少次能让a变成b

思路：先计算出来a和b的差值呗 cha=b-a
如果cha==0 就输出0
如果cha>0 说明b比a大 a要加上去 又因为加只能加奇数 我们判断差是奇数还是偶数 是奇数就输出1 偶数就输出2 因为偶数=奇数+奇数
如果cha<0 说明b比a小 a要往下减 还是看差的奇偶 偶数就输出1 奇数就输出2

#include
using namespace std;
typedef long long ll;
#define me(a,x) memset(a,x,sizeof a)
#define rep(i,x,n) for(int i=x;i
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define all(x) (x).begin(), (x).end()
#define pb(a) push_back(a)
#define paii pair
#define pali pair
#define pail pair
#define pall pair
#define fi first
#define se second
int main()
{
   int t;cin>>t;
   while(t--){
      int a,b;
      cin>>a>>b;
      int cha=b-a;
      if(cha==0) puts("0");
      else if(cha>0){
         if(cha&1) puts("1");
         else puts("2");
      }
      else{
        if(cha&1) puts("2");
         else puts("1");
      }

   }
   return 0;
}