Description
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].
问题描述
一步为将(x, y)转换为(x + y, y)或者(x, x + y)
给定起始点(sx, sy)和终点(tx, ty),如果经过一系列的移动能从起点到终点，返回true，否则返回false
问题分析
从(x, y)出发可以到达的点的可能性非常多，而从(tx, ty)倒着推只有一条路
即，若tx > ty, tx = tx % ty,否则，ty = ty % tx
举个例子
tx = 5, ty = 7
由于tx < ty,上一轮肯定是(x, x + y),因此ty = ty %tx = 2
依次类推，可以得到初始点为(1, 2)
思路
当tx > sx 且 ty > sy时，将tx和ty倒推
若tx = sx 且 (ty - sy) % sx == 0 或者 ty == sy 且 (tx - sx) % sy ==0 .返回true
解法
class Solution {
public boolean reachingPoints(int sx, int sy, int tx, int ty) {
while(tx > sx && ty > sy){
if(ty < tx) tx %= ty;
else ty %= tx;
}
if(sx == tx){
return (ty - sy) % sx == 0;
}else if(sy == ty){
return (tx - sx) % sy == 0;
}
return false;
}
}
1 //C++
2 class Solution {
3 public:
4 bool reachingPoints(int sx, int sy, int tx, int ty) {
5 if(tx==sx&&ty==sy)
6 return true;
7 if(tx==ty||tx==0||ty==0)
8 return false;
9 if(ty>tx)
10 return reachingPoints(sx, sy, tx, ty-max((ty-sy)/tx,1)*tx);
11 if(tx>ty)
12 return reachingPoints(sx, sy, tx-max((tx-sx)/ty,1)*ty, ty);
13 return false;
14 }
15 };
1 #include <stdio.h>
2 #include <math.h>
3
4
5 #if 1
6 #define MY_print_ln_E(fmt, arg...) { if (1 <= g_log_level) {printf("(%s|%d)" fmt"\r\n", __func__, __LINE__, ##arg);} }
7 #define MY_print_ln_W(fmt, arg...) { if (2 <= g_log_level) {printf("(%s|%d)" fmt"\r\n", __func__, __LINE__, ##arg);} }
8 #define MY_print_ln_I(fmt, arg...) { if (3 <= g_log_level) {printf("(%s|%d)" fmt"\r\n", __func__, __LINE__, ##arg);} }
9 #define MY_print_ln_D(fmt, arg...) { if (4 <= g_log_level) {printf("(%s|%d)" fmt"\r\n", __func__, __LINE__, ##arg);} }
10 #else
11 #define MY_print_ln_E(fmt, arg...) { }
12 #define MY_print_ln_W(fmt, arg...) { }
13 #define MY_print_ln_I(fmt, arg...) { }
14 #define MY_print_ln_D(fmt, arg...) { }
15 #endif
16 #define false (0)
17 #define true (1)
18
19 static int g_log_level = 3;
20
21 int max(int a, int b)
22 {
23 return a > b ? a : b;
24 }
25
26 /**********************************************************************************************
27 坐标点移动能力测试
28
29 **** 坐标在第一象限移动 ****
30
31 日期: 2020.09.28 21:00:00
32 作者: LiuYan
33
34 参数:
35 sx, 起始点横坐标
36 sy, 起始点纵坐标
37 tx, 起始点横坐标
38 ty, 起始点纵坐标
39
40 返回值: 1能移动到目标坐标, 0不能移动到目标坐标
41
42 **********************************************************************************************/
43 int move_from_to(int sx, int sy, int tx, int ty)
44 {
45 int tmp_x = 0;
46 int tmp_y = 0;
47 int move_who = 0; // 0为x, 1为y
48 int n = 0;
49
50
51 MY_print_ln_D("judge from=(%d, %d), to=(%d, %d)\r\n", sx, sy, tx, ty);
52
53 for (tmp_x = tx, tmp_y = ty; !(tmp_x < sx && tmp_y < sy); )
54 {
55
56 if (tmp_x == sx && tmp_y == sy)
57 {
58 MY_print_ln_I("move (%d, %d)->(%d, %d), Success", sx, sy, tx, ty);
59 return true;
60 }
61
62 if (tmp_x == tmp_y || 0 == tmp_x || 0 == tmp_y)
63 {
64 MY_print_ln_I("move (%d, %d)->(%d, %d), Failed, last_point=(%d, %d)", sx, sy, tx, ty, tmp_x, tmp_y);
65 return false;
66 }
67
68 move_who = (tmp_y > tmp_x) ? 1 : 0;
69
70 if (0 == move_who) // 移动x
71 {
72 n = (tmp_x-sx)/tmp_y;
73 n = n < 1 ? 1 : n;
74
75 tmp_x -= n * tmp_y;
76 }
77 else // 移动y
78 {
79 n = (tmp_y-sy)/tmp_x;
80 n = n < 1 ? 1 : n;
81
82 tmp_y -= n * tmp_x;
83 }
84
85 MY_print_ln_D("move_who=%d, (%d, %d)", move_who, tmp_x, tmp_y);
86
87 }
88
89 MY_print_ln_I("move (%d, %d)->(%d, %d), Failed, last_point=(%d, %d)", sx, sy, tx, ty, tmp_x, tmp_y);
90 return false;
91 }
92
93 int main(int argc, char *argv[])
94 {
95 // 成功测试用例
96 MY_print_ln_I("%d", move_from_to(1, 2, 5, 2));
97 MY_print_ln_I("%d", move_from_to(2, 3, 2, 9));
98 MY_print_ln_I("%d", move_from_to(1, 2, 3, 5));
99 MY_print_ln_I("%d", move_from_to(1, 2, 3, 8));
100 MY_print_ln_I("%d", move_from_to(1, 2, 11, 8));
101 MY_print_ln_I("%d", move_from_to(1, 2, 19, 8));
102 MY_print_ln_I("%d", move_from_to(1, 2, 27, 8));
103
104 // 失败测试用例
105 MY_print_ln_I("%d", move_from_to(1, 2, 27, 9));
106 return 0;
107 }
