1237. 找出给定方程的正整数解

给出一个函数  f(x, y) 和一个目标结果 z，请你计算方程 f(x,y) == z 所有可能的正整数 数对 x 和 y。

给定函数是严格单调的，也就是说：

f(x, y) < f(x + 1, y)
f(x, y) < f(x, y + 1)
函数接口定义如下：

interface CustomFunction {
public:
  // Returns positive integer f(x, y) for any given positive integer x and y.
  int f(int x, int y);
};
如果你想自定义测试，你可以输入整数 function_id 和一个目标结果 z 作为输入，其中 function_id 表示一个隐藏函数列表中的一个函数编号，题目只会告诉你列表中的 2 个函数。  

你可以将满足条件的 结果数对 按任意顺序返回。

示例 1：

输入：function_id = 1, z = 5
输出：[[1,4],[2,3],[3,2],[4,1]]
解释：function_id = 1 表示 f(x, y) = x + y
示例 2：

输入：function_id = 2, z = 5
输出：[[1,5],[5,1]]
解释：function_id = 2 表示 f(x, y) = x * y
 

提示：

1 <= function_id <= 9
1 <= z <= 100
题目保证 f(x, y) == z 的解处于 1 <= x, y <= 1000 的范围内。
在 1 <= x, y <= 1000 的前提下，题目保证 f(x, y) 是一个 32 位有符号整数。

来源：力扣（LeetCode）
链接：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

 利用单调性

"""
   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