AtCoder Grand Contest 008 A

Problem Statement

Snuke has a calculator. It has a display and two buttons.

Initially, the display shows an integer x. Snuke wants to change this value into another integer y, by pressing the following two buttons some number of times in arbitrary order:

• Button A: When pressed, the value on the display is incremented by 1.
• Button B: When pressed, the sign of the value on the display is reversed.

Find the minimum number of times Snuke needs to press the buttons to achieve his objective. It can be shown that the objective is always achievable regardless of the values of the integers x and y.

Constraints

• x and y are integers.
• |x|,|y|≤109
• x and y are different.

---

Input

The input is given from Standard Input in the following format:

x y

Output

Print the minimum number of times Snuke needs to press the buttons to achieve his objective.

---

Sample Input 1

Copy

10 20

Sample Output 1

Copy

10

Press button A ten times.

---

Sample Input 2

Copy

10 -10

Sample Output 2

Copy

1

Press button B once.

---

Sample Input 3

Copy

-10 -20

Sample Output 3

Copy

12

Press the buttons as follows:

• Press button B once.
• Press button A ten times.
• Press button B once.

题意：告诉我们两个数字，两种操作，一种是+1，一种是改变正负，问最小进行几次操作可以将x变成y

解法：讨论讨论讨论。。。。（注意等于0的情况）

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
    ll n,m;
    cin>>n>>m;
//   cout<<min(abs(m-n),min(abs(-1*n-m)+1,min(abs(-1*n+m)+1,min(abs(-1*n+m)+2,abs(-1*n-m)+2))))<<endl;
   // if(n==m) continue;
   // cout<<n<<" "<<m<<" ";
      if(n<m)
            {
                ll pos=n*(-);
                if(n>=&&m>=)
                {
                    //3 5
                    cout<<m-n<<endl;
                }
                else if(n<=&&m>=)
                {
                    //-3 5
                    //3 5
                    //-3 2
                    //-3+1=-2 2
                    if(pos<=m)
                    {
                        if(n!=)
                        {
                            cout<<m-pos+<<endl;
                        }
                        else
                        {
                            cout<<m-pos<<endl;
                        }
                    }
                    else
                    {
                        // cout<<"A"<<endl;
                        //ll ans=pos-m;
                        if(m==)
                        {
                            cout<<pos<<endl;
                        }
                        else
                            cout<<pos-m+<<endl;
                    }
                }
                else if(n<=&&m<=)
                {
                    //-3 -2
                    cout<<pos+m<<endl;
                }
            }
            else
            {
                if(n>=&&m>=)
                {
                    //4 2
                    if(m!=)
                    {
                        cout<<n-m+<<endl;
                    }
                    else
                    {
                        cout<<n+m+<<endl;
                    }
                }
                else if(n>=&&m<=)
                {
                    ll pos=-*m;
                    if(n==)
                    {
                        cout<<-*m+<<endl;
                    }
                    else
                    {
                        if(n<=pos)
                        {
                            cout<<pos-n+<<endl;
                        }
                        else
                        {
                            cout<<n+m+<<endl;
                        }
                    }
                }
                else
                {
                    if(n==)
                    {
                        cout<<(-)*m+<<endl;
                    }
                    else
                    {
                        cout<<(-)*m-(-)*n+<<endl;
                    }
                }
            }
    return ;
}