Educational Codeforces Round 125 (Rated for Div. 2) ABCD

我服了。

A. Integer Moves

一共三种情况：

• 当前点已经为终点
• 两点之间直线长度为整数，走欧拉距离
• 两点之间直线长度不为整数，走曼哈顿距离

#include 
#define int long long
using namespace std;

const int N = 1e5 + 10;


inline void solve(){
    int x, y; std::cin >> x >> y;
    if(x == 0 && y == 0){ cout  str;
    int ans = 0, cnt = 0;
    for(int l = 0, r = 1; ; ){
        if(r >= n){ cout  atk[cc]) atk[cc] = hh;
    }
    //for(int i = 1; i > b;
        int l = 1, r = C;
        while (l > 1;
            (maxx[mid] > a * b) ? (r = mid - 1) : (l = mid + 1);
        }
        if (l == C + 1) cout