c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)

talk is easy, here is the code

leetcode 1237. 找出给定方程的正整数解链接

解法一 ：暴力求解

    vector<vector<int>> findSolution(CustomFunction &customfunction, int z) {
        vector<vector<int>> ans;
        for (int x = 0; x <= 1000; x++) {
            for (int y = 1; y <= 1000; y++) {
                if (customfunction.f(x, y) == z) {
                    ans.push_back({x, y});
                }
            }
        }
        return ans;
    }

解法二： 二分法

    vector<vector<int>> findSolution(CustomFunction &customfunction, int z) {
        vector<vector<int>> ans;
        struct Node {
            int left, right, top, bottom;
        };

        stack<Node> st;
        st.push({1, 1000, 1, 1000});
        while (!st.empty()) {
            auto node = st.top();
            st.pop();
            if (customfunction.f(node.left, node.top) > z || customfunction.f(node.right, node.bottom) < z) {
                continue;
            }
            int x = (node.right + node.left) / 2;
            int y = (node.bottom + node.top) / 2;
            int v = customfunction.f(x, y);
            if (v < z) {
                if (x + 1 <= node.right) {
                    st.push({x + 1, node.right, node.top, node.bottom});
                }
                if (y + 1 <= node.bottom) {
                    st.push({node.left, x, y + 1, node.bottom});
                }
            } else if (v > z) {
                if (node.left + 1 <= x) {
                    st.push({node.left, x - 1, node.top, node.bottom});
                }
                if (node.top + 1 <= y) {
                    st.push({x, node.right, node.top, y - 1});
                }
            } else {
                ans.push_back({x, y});
                if (x + 1 <= node.right && node.top + 1 <= y) {
                    st.push({x + 1, node.right, node.top, y - 1});
                }
                if (node.left + 1 <= x && y + 1 <= node.bottom) {
                    st.push({node.left, x - 1, y + 1, node.bottom});
                }
            }
        }
        return ans;
    }

解法三： 从左下角往上搜索

    vector<vector<int>> findSolution3(CustomFunction &customfunction, int z) {
        vector<vector<int>> ans;
        int x = 1, y = 1000;
        while (x <= 1000 && y >= 1) {
            int v = customfunction.f(x, y);
            if (v > z) {
                y--;
            } else if (v < z) {
                x++;
            } else {
                ans.push_back({x, y});
                x++;
                y--;
            }
        }
      

解法四: 有限状态机

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

class CustomFunction {
public:
    // Returns f(x, y) for any given positive integers x and y.
    // Note that f(x, y) is increasing with respect to both x and y.
    // i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
    int f(int x, int y) {
        return x * y;
    }
};

enum Direction {
    UP = 0, RIGHT = 2
};

class MoveForward {
public:
    Direction d = UP;

    Direction getDirection() const {
        return d;
    }

    void ChangeDirection() {
        if (d == UP) {
            d = RIGHT;
        } else {
            d = UP;
        }
    }
};

class State {
public:
    State() {
        x = 1;
        y = 1000;
    }

    int x, y;
    MoveForward d;
};

static bool ok(State &s) {
    return s.x <= 1000 && s.y >= 1;
}

// 每次前进1
class NormalStrategy {
public:
    virtual void changeState(State &next) {
        if (next.d.getDirection() == UP) {
            next.y--;
        } else {
            next.x++;
        }
    }
    
    bool tryFit(CustomFunction &customfunction, State &s, int z) {
        State next = s;
        changeState(next);
        // 是否是正常状态
        if (!ok(next)) {
            return false;
        }
        int v = customfunction.f(next.x, next.y);
        do {
            if (v == z) {
                break;
            }
            if (next.d.getDirection() == UP) {
                if (v < z) {
                    return false;
                }
                break;
            }
            if (next.d.getDirection() == RIGHT) {
                if (v > z) {
                    return false;
                }
            }
        } while (false);
        // 更新状态
        s = next;
        return true;
    }
};

// 每次前进一半
class QuarterStrategy : public NormalStrategy {
public:
    void changeState(State &next) override {
        if (next.d.getDirection() == UP) {
            next.y /= 4;
            return;
        } else {
            next.x = (next.x + 3000) / 4;
        }
    }
};

// 每次前进一半
class HalfStrategy : public NormalStrategy {
public:
    void changeState(State &next) override {
        if (next.d.getDirection() == UP) {
            next.y /= 2;
            return;
        } else {
            next.x = (next.x + 1000) / 2;
        }
    }
};

// 每次前进四分之一
class HalfHalfStrategy : public NormalStrategy {
public:
    void changeState(State &next) override {
        if (next.d.getDirection() == UP) {
            next.y = next.y * 3 / 4;
            return;
        } else {
            next.x = (3000 + next.x) / 4;
        }
    }
};

// 每次前进10
class TenStrategy : public NormalStrategy {
public:
    void changeState(State &next) override {
        if (next.d.getDirection() == UP) {
            next.y -= 10;
            return;
        } else {
            next.x += 10;
        }
    }
};

class StateStrategy {
public:
    int up_index = 0;
    int right_index = 0;
    NormalStrategy *strategy[5] = {new QuarterStrategy, new HalfStrategy, new HalfHalfStrategy, new TenStrategy, new NormalStrategy};

    void try_most(CustomFunction &customfunctionm, State &s, int z, int &index) {
        // 尝试尽最大可能前进
        // 最开始急剧的收缩 每次减半
        // 后面每次减少 1 / 4
        // 每次减少 10
        while (index + 2 <= size(strategy)) {
            if (strategy[index]->tryFit(customfunctionm, s, z)) {
                return;
            }
            index++;
        }
        // 委屈求全前进, 每次减少1
        strategy[index]->changeState(s);
    }

    void try_most_right(CustomFunction &customfunctionm, State &s, int z) {
        try_most(customfunctionm, s, z, right_index);
    }

    void try_most_up(CustomFunction &customfunctionm, State &s, int z) {
        try_most(customfunctionm, s, z, up_index);
    }
};

class Saver {
public:
    vector<vector<int>> data;

    virtual void push(int x, int y) {
        data.push_back({x, y});
    }
};

class StateMachine {
public:
    State &s;
    CustomFunction &customfunction;
    void *data;
    StateStrategy ss;

    StateMachine(CustomFunction &customfunction, State &s, void *data) : customfunction(customfunction), s(s),
                                                                         data(data) {}
    void equal_next() {
        s.x++;
        s.y--;
    }

    void equal_state() {
        ((Saver *) (data))->push(s.x, s.y);
        equal_next();
    }

    void nextState(int z) {
        int v = customfunction.f(s.x, s.y);
        if (v == z) {
            equal_state();
            return;
        }
        bool one = v > z;
        auto d = s.d.getDirection();
        // 是否需要调整运动方向
        if (one + d == 0 || one + d == 3) {
            s.d.ChangeDirection();
        }
        if (s.d.getDirection() == UP) {
            ss.try_most_up(customfunction, s, z);
        } else {
            ss.try_most_right(customfunction, s, z);
        }
    }

    void run(int z) {
        while (ok(s)) {
            nextState(z);
        }
    }
};

class Solution {
public:
    vector<vector<int>> findSolution(CustomFunction &customfunction, int z) {
        State s;
        Saver sa;
        StateMachine machine(customfunction, s, &sa);
        machine.run(z);
        return sa.data;
    }
};

int main() {
    CustomFunction x;
    auto data = Solution{}.findSolution(x, 5);
    for (auto &vec: data) {
        for (auto v: vec) {
            cout << v << "\t";
        }
        cout << endl;
    }
}

c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)
c++1237. 找出给定方程的正整数解，四种解法(二分+有限状态机)