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" ", __func__, __LINE__, ##arg);} } 7 #define MY_print_ln_W(fmt, arg...) { if (2 <= g_log_level) {printf("(%s|%d)" fmt" ", __func__, __LINE__, ##arg);} } 8 #define MY_print_ln_I(fmt, arg...) { if (3 <= g_log_level) {printf("(%s|%d)" fmt" ", __func__, __LINE__, ##arg);} } 9 #define MY_print_ln_D(fmt, arg...) { if (4 <= g_log_level) {printf("(%s|%d)" fmt" ", __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) ", 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 }