給出一個函式f(x, y)和一個目標結果z，請你計算方程f(x,y) == z所有可能的正整數 數對x 和 y。

給定函式是嚴格單調的，也就是說：

f(x, y) < f(x + 1, y)
f(x, y) < f(x, y + 1)
函式介面定義如下：

interface CustomFunction {
public:
// Returns positive integer f(x, y) for any given positive integer x and y.
int f(int x, int y);
};
如果你想自定義測試，你可以輸入整數function_id和一個目標結果z作為輸入，其中function_id表示一個隱藏函式列表中的一個函式編號，題目只會告訴你列表中的 2 個函式。

你可以將滿足條件的 結果數對 按任意順序返回。

示例 1：

輸入：function_id = 1, z = 5
輸出：[[1,4],[2,3],[3,2],[4,1]]
解釋：function_id = 1 表示 f(x, y) = x + y
示例 2：

輸入：function_id = 2, z = 5
輸出：[[1,5],[5,1]]
解釋：function_id = 2 表示 f(x, y) = x * y

提示：

1 <= function_id <= 9
1 <= z <= 100
題目保證f(x, y) == z的解處於1 <= x, y <= 1000的範圍內。
在 1 <= x, y <= 1000的前提下，題目保證f(x, y)是一個32 位有符號整數。

來源：力扣（LeetCode）
連結：https://leetcode-cn.com/problems/find-positive-integer-solution-for-a-given-equation

暴力

"""
   This is the custom function interface.
   You should not implement it, or speculate about its implementation
   class CustomFunction:
       # 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)
       def f(self, x, y):
  
 
"""

class Solution:
    def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
        res=[]
        for i in range(1,z+1):
            for j in range(1,z+1):
                if customfunction.f(i,j)==z:
                    res.append([i,j])
        return res

二分

"""
   This is the custom function interface.
   You should not implement it, or speculate about its implementation
   class CustomFunction:
       # 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)
       def f(self, x, y):
  
"""

class Solution:
    def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
        res=[]
        for i in range(1,z+1):
            l=1
            r=z
            while l<=r:
                mid=(l+r)>>1
                fun=customfunction.f(i,mid)
                if fun==z:
                    res.append([i,mid])
                    break
                elif fun<z:
                    l=mid+1
                else:
                    r=mid-1
        return res

利用單調性

"""
   This is the custom function interface.
   You should not implement it, or speculate about its implementation
   class CustomFunction:
       # 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)
       def f(self, x, y):
  
"""

class Solution:
    def findSolution(self, customfunction: 'CustomFunction', z: int) -> List[List[int]]:
        y=z
        for x in range(1,z+1):
            while y:
                if customfunction.f(x,y)<=z:
                    if customfunction.f(x,y)==z:
                        yield [x,y]
                    break
                y-=1