Add Odd or Subtract Even
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 xx (x>0x>0) and replace aa with a xa x;
- choose any positive even integer yy (y>0y>0) and replace aa with a−ya−y.
You can perform as many such operations as you want. You can choose the same numbers xx and yy in different moves.
Your task is to find the minimum number of moves required to obtain bb from aa. It is guaranteed that you can always obtain bb from aa.
You have to answer tt independent test cases.
Input
The first line of the input contains one integer tt (1≤t≤1041≤t≤104) — the number of test cases.
Then tt test cases follow. Each test case is given as two space-separated integers aa and bb (1≤a,b≤1091≤a,b≤109).
Output
For each test case, print the answer — the minimum number of moves required to obtain bb from aa if you can perform any number of moves described in the problem statement. It is guaranteed that you can always obtain bb from aa.
Example
input
代码语言：javascript复制5
2 3
10 10
2 4
7 4
9 3output
代码语言：javascript复制1
0
2
2
1Note
In the first test case, you can just add 11.
In the second test case, you don't need to do anything.
In the third test case, you can add 11 two times.
In the fourth test case, you can subtract 44 and add 11.
In the fifth test case, you can just subtract 66.
手速题：如果a比b小，差值是奇数 1次，否则两次，b比a小，差值是偶数 1次，否则两次
代码语言：javascript复制#include<bits/stdc .h>
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template <typename T> inline void read(T& x)
{
x=0;char ch=getchar();ll f=1;
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=(x<<3) (x<<1) (ch^48);ch=getchar();}x*=f;
}
inline ll query(ll x){ll res=0;while(x){res =sz[x];x-=lb(x);}return res;}
inline void add(ll x,ll val){while(x<=n){sz[x] =val;x =lb(x);}}//第x个加上val
ll x,y,t;
int main()
{
cin>>t;
for(rg i=1;i<=t;i )
{
cin>>x>>y;
if(x==y)
{
cout<<0<<endl;
continue;
}
if(x>y)
{
if((x-y)%2)
{
cout<<2<<endl;
}
else cout<<1<<endl;
}
else
{
if((y-x)%2)
{
cout<<1<<endl;
}
else cout<<2<<endl;
}
}
//while(1)getchar();
return 0;
}
