> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/numbers/780.reaching-points.md).

# 780.Reaching-Points

## 780. Reaching Points

## 题目地址

<https://leetcode.com/problems/reaching-points/>

## 题目描述

```
A move consists of taking a point (x, y) and transforming it to either (x, x+y) or (x+y, y).

Given a starting point (sx, sy) and a target point (tx, ty), return True if and only if a sequence of moves exists to transform the point (sx, sy) to (tx, ty). Otherwise, return False.

Examples:
Input: sx = 1, sy = 1, tx = 3, ty = 5
Output: True
Explanation:
One series of moves that transforms the starting point to the target is:
(1, 1) -> (1, 2)
(1, 2) -> (3, 2)
(3, 2) -> (3, 5)

Input: sx = 1, sy = 1, tx = 2, ty = 2
Output: False

Input: sx = 1, sy = 1, tx = 1, ty = 1
Output: True

Note:
sx, sy, tx, ty will all be integers in the range [1, 10^9].
```

## 代码

### Approach #1

**Basic idea:** If we start from `sx,sy`, it will be hard to find `tx, ty`. If we start from `tx,ty`, we can find only one path to go back to `sx, sy`. I cut down one by one at first and I got TLE. So I came up with remainder.

**First line:** if 2 target points are still bigger than 2 starting point, we reduce target points. **Second line:** check if we reduce target points to (x, y+kx) or (x+ky, y)

**Time complexity** I will say `O(logN)` where `N = max(tx,ty)`.

```java
class Solution {
  while (sx < tx && sy < ty) {
    if (tx < ty) {
      ty %= tx;
    } else {
      tx %= ty;
    }
  }
  return sx == tx && sy <= ty && (ty - sy) % sx == 0 ||
             sy == ty && sx <= tx && (tx - sx) % sy == 0;
}
```