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.
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.
输入
The input is given from Standard Input in the following format:
x y
x y
输出
Print the minimum number of times Snuke needs to press the buttons to achieve his objective.
水题，思路清晰一般不会错。
题意；给你x，y两个数，有两种操作：操作（1）x的数值+1，操作（2）x=-x；问你最少走多少步。
思路比较简单，就不写了
#include<stdio.h>
#include<math.h>
#include<string.h>
using
namespace
std;
#define ll long long
ll min(ll a,ll b ) {
return
a<b?a:b;}
ll max(ll a,ll b) {
return
a>b?a:b;}
ll minn(ll a,ll b,ll c){
return
min(min(a,b),min(b,c));}
ll maxn(ll a,ll b,ll c){
return
max(max(a,b),max(b,c));}
int
main()
{
int
x,y;
scanf
(
"%d%d"
,&x,&y);
if
(x==-y)
printf
(
"1\n"
);
else
if
(x==0) {
if
(y>0)
printf
(
"%d\n"
,y);
else
printf
(
"%d\n"
,-y+1);
}
else
if
(y==0)
{
if
(x<0)
printf
(
"%d\n"
,-x);
else
printf
(
"%d\n"
,x+1);
}
else
if
(0<x&&x<y)
printf
(
"%d\n"
,y-x);
else
if
(0<y&&y<x)
printf
(
"%d\n"
,x-y+2);
else
if
(x>0&&y<0)
{
if
(x>-y)
printf
(
"%d\n"
,x+y+1);
else
if
(x<-y)
printf
(
"%d\n"
,-y-x+1);
}
else
if
(x<0&&y>0)
{
if
(-x<y)
printf
(
"%d\n"
,y+x+1);
else
printf
(
"%d\n"
,-x-y+1);
}
else
if
(x<y&&y<0)
printf
(
"%d\n"
,y-x);
else
if
(y<x&&x<0)
printf
(
"%d\n"
,x-y+2);
return
0;
}
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