One of our delivery robots is malfunctioning! The job of the robot is simple; it should follow a list of instructions in order to reach a target destination. The list of instructions is originally correct to get the robot to the target. However, something is going wrong as we upload the instructions into the robot’s memory. During the upload, one random instruction from the list takes on a different value than intended. Yes, there is always a single bad instruction in the robot’s memory and it always results in the robot arriving at an incorrect destination as it finishes executing the list of instructions.
The robot can execute the instructions “Left”, “Right”, and “Forward”. The “Left” and “Right” instructions do not result in spatial movement but result in a 9090-degree turn in the corresponding direction. “Forward” is the only instruction that results in spatial movement, causing the robot to move one unit in the direction it is facing. The robot always starts at the origin (0,0)(0,0) of a grid and faces north along the positive y-axis.
Given the coordinates of the target destination and the list of instructions that the robot has in its memory, you are to identify a correction to the instructions to help the robot reach the proper destination.
Input
The first line of the input contains the xx and yy integer coordinates of the target destination, where ?50≤x≤50?50≤x≤50 and ?50≤y≤50?50≤y≤50. The following line contains an integer nn representing the number of instructions in the list, where 1≤n≤501≤n≤50. The remaining nn lines each contain a single instruction. These instructions may be: “Left”, “Forward”, or “Right”.
Output
Identify how to correct the robot’s instructions by printing the line number (starting at 11) of an incorrect input instruction, followed by an instruction substitution that would make the robot reach the target destination. If there are multiple ways to fix the instructions, report the fix that occurs for the earliest line number in the sequence of instructions. There is always exactly one unique earliest fix.
| Sample Input 1 | Sample Output 1 |
|---|---|
3 2 11 Forward Right Forward Forward Left Forward Forward Left Forward Right Forward |
8 Right |
| Sample Input 2 | Sample Output 2 |
|---|---|
-1 1 3 Right Left Forward |
1 Forward |
题解：给一个指定的点，给机器人下指令，前进、向右或者向左转，然后机器人出了故障，中间有一个命令出错了（可能有多种解决方案，输出最早的），找到第几个出错，并输出正确指令。
1 #include<iostream> 2 #include<cstdlib> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 #include<algorithm> 7 #include<vector> 8 #include<string> 9 using namespace std; 10 int x, y; 11 int n; 12 char ins[500][10]; 13 int map[1][2] = {0}; 14 int go[4][2] = { {1,0},{0,1},{-1,0},{0,-1} }; 15 int main() 16 { 17 int turn = 0; 18 scanf("%d %d", &x, &y); 19 scanf("%d", &n); 20 for (int i = 1; i <= n; i++) 21 { 22 scanf("%s", ins[i]); 23 } 24 for (int i = 1; i <= n; i++) 25 { 26 if (strcmp(ins[i], "Forward") == 0) 27 { 28 strcpy(ins[i],"Right"); 29 turn = 0; 30 for (int j = 1; j <= n; j++) 31 { 32 if (strcmp(ins[j], "Forward") == 0) 33 { 34 map[0][0] += go[turn][0]; //y 35 map[0][1] += go[turn][1]; //x 36 } 37 else if (strcmp(ins[j], "Right") == 0) 38 { 39 turn++; 40 if (turn > 3) turn = 0; 41 } 42 else 43 { 44 turn--; 45 if (turn < 0) turn = 3; 46 } 47 } 48 if (map[0][0] == y&&map[0][1] == x) 49 { 50 printf("%d %s\n", i, ins[i]); 51 break; 52 } 53 map[0][0] = 0; 54 map[0][1] = 0; 55 strcpy(ins[i], "Left"); 56 turn = 0; 57 for (int j = 1; j <= n; j++) 58 { 59 if (strcmp(ins[j], "Forward") == 0) 60 { 61 map[0][0] += go[turn][0]; //y 62 map[0][1] += go[turn][1]; //x 63 } 64 else if (strcmp(ins[j], "Right") == 0) 65 { 66 turn++; 67 if (turn > 3) turn = 0; 68 } 69 else 70 { 71 turn--; 72 if (turn < 0) turn = 3; 73 } 74 } 75 if (map[0][0] == y&&map[0][1] == x) 76 { 77 printf("%d %s\n", i, ins[i]); 78 break; 79 } 80 strcpy(ins[i], "Forward"); 81 map[0][0] = 0; 82 map[0][1] = 0; 83 } 84 else if (strcmp(ins[i], "Right") == 0) 85 { 86 87 strcpy(ins[i], "Forward"); 88 turn = 0; 89 for (int j = 1; j <= n; j++) 90 { 91 if (strcmp(ins[j], "Forward") == 0) 92 { 93 map[0][0] += go[turn][0]; //y 94 map[0][1] += go[turn][1]; //x 95 } 96 else if (strcmp(ins[j], "Right") == 0) 97 { 98 turn++; 99 if (turn > 3) turn = 0; 100 } 101 else 102 { 103 turn--; 104 if (turn < 0) turn = 3; 105 } 106 } 107 if (map[0][0] == y&&map[0][1] == x) 108 { 109 printf("%d %s\n", i, ins[i]); 110 break; 111 } 112 map[0][0] = 0; 113 map[0][1] = 0; 114 strcpy(ins[i], "Left"); 115 turn = 0; 116 for (int j = 1; j <= n; j++) 117 { 118 if (strcmp(ins[j], "Forward") == 0) 119 { 120 map[0][0] += go[turn][0]; //y 121 map[0][1] += go[turn][1]; //x 122 } 123 else if (strcmp(ins[j], "Right") == 0) 124 { 125 turn++; 126 if (turn > 3) turn = 0; 127 } 128 else 129 { 130 turn--; 131 if (turn < 0) turn = 3; 132 } 133 } 134 if (map[0][0] == y&&map[0][1] == x) 135 { 136 printf("%d %s\n", i, ins[i]); 137 break; 138 } 139 strcpy(ins[i], "Right"); 140 map[0][0] = 0; 141 map[0][1] = 0; 142 } 143 else 144 { 145 146 strcpy(ins[i], "Right"); 147 turn = 0; 148 for (int j = 1; j <= n; j++) 149 { 150 if (strcmp(ins[j], "Forward") == 0) 151 { 152 map[0][0] += go[turn][0]; //y 153 map[0][1] += go[turn][1]; //x 154 } 155 else if (strcmp(ins[j], "Right") == 0) 156 { 157 turn++; 158 if (turn > 3) turn = 0; 159 } 160 else 161 { 162 turn--; 163 if (turn < 0) turn = 3; 164 } 165 } 166 if (map[0][0] == y&&map[0][1] == x) 167 { 168 printf("%d %s\n", i, ins[i]); 169 break; 170 } 171 map[0][0] = 0; 172 map[0][1] = 0; 173 strcpy(ins[i], "Forward"); 174 turn = 0; 175 for (int j = 1; j <= n; j++) 176 { 177 if (strcmp(ins[j], "Forward") == 0) 178 { 179 map[0][0] += go[turn][0]; //y 180 map[0][1] += go[turn][1]; //x 181 } 182 else if (strcmp(ins[j], "Right") == 0) 183 { 184 turn++; 185 if (turn > 3) turn = 0; 186 } 187 else 188 { 189 turn--; 190 if (turn < 0) turn = 3; 191 } 192 } 193 if (map[0][0] == y&&map[0][1] == x) 194 { 195 printf("%d %s\n", i, ins[i]); 196 break; 197 } 198 strcpy(ins[i], "Left"); 199 map[0][0] = 0; 200 map[0][1] = 0; 201 } 202 203 } 204 return 0; 205 }