7-116 3207 数字阶梯

这题很有意思啊，写个解析。
观察数字坐标的移动方式，不难发现所有的移动以 ↗↘↗↖ 为一组不断循环，且每次移动后对应坐标的点变为上个点 \(+1\) 。
那么代码就很好模拟了，我们只需要使用 {{1, 1}, {1, -1}, {1, 1}, {-1, 1}} 来模拟每次的移动即可。

#include <algorithm>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <limits>
#include <map>
#include <numeric>
#include <string>
#include <utility>
#include <vector>

using i32 = std::int32_t;
using i64 = std::int64_t;
using u32 = std::uint32_t;
using u64 = std::uint64_t;
using pii = std::pair<i32, i32>;
using pll = std::pair<i64, i64>;
using vi = std::vector<i32>;
using vll = std::vector<i64>;
using vpii = std::vector<pii>;
using vpll = std::vector<pll>;

void solve();

int main()
{
    std::cin.tie(nullptr)->sync_with_stdio(false);
    solve();
}

using move = pii;
using point = pii;

class MoveManager
{
  public:
    move get_move()
    {
        ++index;
        if (index <= 3)
        {
            return moves[index];
        }
        index = 0;
        return moves[0];
    }

  private:
    i32 index{-1};
    const std::vector<move> moves{{1, 1}, {1, -1}, {1, 1}, {-1, 1}};
};

void solve()
{

    std::map<point, i32> map;
    point current_point{0, 0};
    i32 current_value{0};
    MoveManager movemanager;
    while (current_point.first < 5000 || current_point.second < 5000)
    {
        map[current_point] = current_value;
        auto get = movemanager.get_move();
        current_point.first += get.first;
        current_point.second += get.second;
        ++current_value;
    }
    i32 t;
    std::cin >> t;
    bool endl = false;
    for (i32 i = 1; i <= t; ++i)
    {
        if (endl)
        {
            std::cout << '\n';
        }
        i32 x, y;
        std::cin >> x >> y;
        if (x == 0 && y == 0)
        {
            std::cout << 0;
        }
        else
        {
            if (map[{x, y}] > 0)
            {
                std::cout << map[{x, y}];
            }
            else
            {
                std::cout << "No Number";
            }
        }
        endl = true;
    }
}