leetcode 春季赛遇到了

来补一发题解

思路：
    计算内侧正方形有多少个
    cnt = max(abs(xPos), abs(yPos)) - 1;
    cnt = max(0ll, cnt);

    运用等差数列求和
    outside = t * 2;
    inside = 2;
    如果inside > outside： now = 0

    然后计算当前边长
    util = t + 1 << 1

    分四种情况去特判

#include <iostream>

using namespace std;

typedef long long LL;

LL xPos, yPos;
LL now;

int main(){

    cin >> xPos >> yPos;

    LL t = max(abs(xPos), abs(yPos)) - 1;
    t = max(0ll, t);
    LL outside = (t * 2);
    LL inside = 2;

    if(outside >= inside) now = (outside + inside) * t / 2ll * 4;

    LL util = (t + 1) << 1;
    if(xPos == t + 1){
        now += util * 2 + abs(yPos - (t + 1));
    }else if(xPos == -(t + 1)){
        now += abs(yPos + (t + 1));
    }else if(yPos == t + 1){
        now += util + abs(xPos + (t + 1));
    }else if(yPos == -(t + 1)){
        now += util * 3 + abs(xPos - (t + 1));
    }

    cout << now << endl;

    return 0;
}