#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const char inst[] = "ABLR";
const int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int main(void)
{
int t = 0;
char s[5][6];
char c;
while ((s[0][0] = getchar()) != 'Z') {
int bi = 0, bj = 0;
for (int i = 0; i < 5; i ++) {
for (int j = 0; j < 5; j ++) {
if (!i && !j) continue;
s[i][j] = getchar();
if (s[i][j] == ' ') {bi = i, bj = j;}
}
getchar();
}
bool legal = true;
while ((c = getchar()) != '0') {
if (legal == false || c == '\n') continue;
int k;
for (k = 0; k < 4; k ++) {
if (c == inst[k]) break;
}
if (k == 4)
legal = false;
else {
int ni = bi+dir[k][0], nj = bj+dir[k][1];
if (0 <= ni && ni < 5 && 0 <= nj && nj < 5) {
swap(s[bi][bj], s[ni][nj]);
bi = ni, bj = nj;
} else
legal = false;
}
}
if (++t > 1) printf("\n");
printf("Puzzle #%d:\n", t);
if (legal == false)
printf("This puzzle has no final configuration.\n");
else {
for (int i = 0; i < 5; i ++) {
for (int j = 0; j < 5; j ++) {
printf("%c%c", s[i][j], j == 4 ? '\n' : ' ');
}
}
}
getchar();
}
return 0;
}
1 # include <iostream>
2 # include <cstdio>
3
4 using namespace std;
5
6 int main(){
7
8 char a[5][7],t;
9 int n,m,i,j,k;
10
11
12
13 int cases = 0;
14 char modol[1001];
15 while(gets(a[0])){
16
17 if(a[0][0]=='Z') break;
18
19 for(i=1;i<5;i++){
20 gets(a[i]);
21 }
22
23
24 //查找空格的位置
25 int x = 0,y = 0;
26 for(i=0;i<5;i++)
27 for(j=0;j<5;j++)
28 if(a[i][j]==' '){
29 x=i,y=j;
30 break;//记录空格这个点
31 }
32
33 //scanf不读取\n和空格
34 int cnt = 0;
35
36
37 while(~scanf("%c",&modol[cnt]))
38 if(modol[cnt]!='0') cnt++;
39 else break;
40 /*
41 char ch;
42 while(scanf("%c",&ch))
43 if(ch=='\n') continue;
44 else if(ch==' ') continue;
45 else if(ch!='0') {
46 modol[cnt]=ch;
47 cnt++;
48 }
49 else break;
50 */
51 //gets与scanf("%c") getcahr 的区别
52 int flag=0, x1 = x , y1=y;
53 modol[cnt] = 0;
54 getchar();
55
56 for(i=0;modol[i];i++){
57 switch(modol[i]){
58 case 'A' : x1 = x - 1;y1=y; break;
59 case 'B' : x1 = x + 1;y1=y; break;
60 case 'L' : x1 = x;y1=y-1; break;
61 case 'R' : x1 = x;y1=y+1; break;
62 }
63
64 if(x1<0||y1<0||x1>4||y1>4){
65 flag = 1;break;
66 }else{
67 a[x][y] = a[x1][y1];
68 a[x1][y1] = ' ';
69 x = x1;
70 y = y1;
71 }
72
73 }
74
75 if(cases++) printf("\n");
76 printf("Puzzle #%d:\n",cases);
77
78 if(flag){
79 printf("This puzzle has no final configuration.\n");
80 }else{
81 for(i=0;i<5;i++){
82 printf("%c",a[i][0]);
83 for(j=1;j<5;j++){
84 printf(" %c",a[i][j]);
85 }
86 printf("\n");
87 }
88
89
90
91 }
92
93
94
95 }
96
97
98 return 0;
99 } 
MDPI工作强度 552人发布
「端到端耗时缩减30-40%」要给出确切数字和绝对值。从1000ms降到600ms是降了40%，从100ms降到60ms也是降了40%，但这两个含义完全不一样。其他也是，涉及到数据，准备好证据，口径统一，面试会问
「熟练」「熟悉」「了解」混在一起用，读起来很乱。而且「了解前端需求」最好改成「具备前后端协作经验」