模拟题 poj 1068、 2632 、 1573、 2993、 2996

poj 1068 Parencodings

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 int vis[100];
 5 int q[100];
 6 void chack(char str[])
 7 {
 8 
 9      int count=0;
10      int i,j,k,d;
11      int len=strlen(str);
12      for(i=0;i<len;i++)
13      {
14          if(str[i]==')'&&!vis[i])
15          {
16              vis[i]=1;
17              for(j=i-1;j>=0;j--)
18              {
19                  if(str[j]=='('&&!vis[j])
20                     {
21                         vis[j]=1;
22                         break;
23                     }
24              }
25              d=0;
26              for(k=j;k<i;k++)
27                {
28                    if(str[k]=='(')
29                     d++;
30                }
31                q[count++]=d;
32          }
33      }
34      for(i=0;i<count-1;i++)
35       printf("%d ",q[i]);
36       printf("%d\n",q[i]);
37 }
38 int main()
39 {
40     int i,j;
41     int s[100];
42     int n,m;
43     char str[100];
44     scanf("%d",&n);
45     while(n--)
46     {
47         int cot=0;
48         int k=0;
49         scanf("%d",&m);
50         for(i=0; i<m; i++)
51             scanf("%d",&s[i]);
52         for(i=0; i<m; i++)
53         {
54             for(j=cot; j<s[i]; j++)
55                 str[k++]='(';
56             str[k++]=')';
57             cot=s[i];
58         }
59         str[k]='\0';
60         memset(vis,0,sizeof(vis));
61         chack(str);
62     }
63     return 0;
64 }

View Code

poj 2632 Crashing Robots

http://poj.org/problem?id=2632

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<stdlib.h>
  4 struct node
  5 {
  6     int x;
  7     int y;
  8     char f;
  9 } ro[101];
 10 struct no
 11 {
 12     int num;
 13     char  zh;
 14     int count;
 15 } z[101];
 16 int n,m;
 17 int a,b;
 18 int i,j,k;
 19 int flag;
 20 void zx()
 21 {
 22     int nu,cot;
 23     char h;
 24     for(i=0; i<b; i++)
 25     {
 26         nu=z[i].num;
 27         h=z[i].zh;
 28         cot=z[i].count;
 29         for(j=0; j<cot; j++)
 30         {
 31             if(h=='F')    //直走
 32             {
 33                 if(ro[nu].f=='E')
 34                     ro[nu].x++;
 35                 else if(ro[nu].f=='W')
 36                     ro[nu].x--;
 37                 else if(ro[nu].f=='N')
 38                     ro[nu].y++;
 39                 else if(ro[nu].f=='S')
 40                     ro[nu].y--;
 41                 for(k=1; k<=a; k++)
 42                 {
 43                     if(k==nu)
 44                         continue;
 45                     if(ro[nu].x==ro[k].x&&ro[nu].y==ro[k].y)
 46                     {
 47                         flag=1;
 48                         printf("Robot %d crashes into robot %d\n",nu,k);
 49                         return;
 50                     }
 51                 }
 52                 if(ro[nu].x>n||ro[nu].x<1||ro[nu].y>m||ro[nu].y<1)
 53                 {
 54                     flag=1;
 55                     printf("Robot %d crashes into the wall\n",nu);
 56                     return;
 57                 }
 58             }
 59             else if(h=='L')// 左转
 60             {
 61                 if(ro[nu].f=='E')
 62                     ro[nu].f='N';
 63                 else if(ro[nu].f=='W')
 64                     ro[nu].f='S';
 65                 else if(ro[nu].f=='N')
 66                     ro[nu].f='W';
 67                 else if(ro[nu].f=='S')
 68                     ro[nu].f='E';
 69             }
 70             else if(h=='R')//右转
 71             {
 72                 if(ro[nu].f=='E')
 73                     ro[nu].f='S';
 74                 else if(ro[nu].f=='W')
 75                     ro[nu].f='N';
 76                 else if(ro[nu].f=='N')
 77                     ro[nu].f='E';
 78                 else if(ro[nu].f=='S')
 79                     ro[nu].f='W';
 80             }
 81         }
 82     }
 83     if(flag==0)
 84     printf("OK\n");
 85 }
 86 int main()
 87 {
 88 
 89     int t;
 90     scanf("%d",&t);
 91     while(t--)
 92     {
 93         scanf("%d %d",&n,&m);
 94         scanf("%d %d",&a,&b);
 95          flag=0;
 96         for(i=1; i<=a; i++)//输入机器人信息
 97             scanf("%d %d %c",&ro[i].x,&ro[i].y,&ro[i].f);
 98         for(i=0; i<b; i++)？//输入命令
 99             scanf("%d %c %d",&z[i].num,&z[i].zh,&z[i].count);
100         zx();//执行命令
101     }
102     return 0;
103 
104 }

View Code

poj 1573 Robot Motion

http://poj.org/problem?id=1573

 1 #include<stdio.h>
 2 #include<string.h>
 3 int main()
 4 {
 5     int row,col,entry;
 6     char s[12][12];
 7     while(~scanf("%d %d %d%*c",&row,&col,&entry))
 8     {
 9         if(!(row ||col || entry))
10           break;
11         memset(s,'O',sizeof(s));
12         int i,j;
13         char ch;
14         for(i=1;i<=row;i++)
15            {
16                for(j=1;j<=col;j++)
17                scanf("%c",&s[i][j]);
18                scanf("%*c",&ch);
19            }
20             int flag[12][12]={0};
21             int count;
22             int r=1;
23             int c=entry;
24             for(count=0; ;count++)
25             {
26                 flag[r][c]++;
27                 if(s[r][c]=='N')
28                     r--;
29                 else if(s[r][c]=='S')
30                     r++;
31                 else if(s[r][c]=='W')
32                     c--;
33                 else if(s[r][c]=='E')
34                     c++;
35                 else if(s[r][c]=='O')
36                 {
37                    printf("%d step(s) to exit\n",count);
38                     break;
39                 }
40              if(flag[r][c]==2)
41                 {
42                     row=r;
43                     col=c;
44                     int flg=1;
45                     for(r=1,c=entry,count=0;;count++)
46                     {
47                         if(r==row && c==col && flg==1)
48                         {
49                             printf("%d step(s) before a loop of ",count);
50                             count=0;
51                             flg++;
52                         }
53                         if(r==row && c==col && count!=0 && flg==2)
54                         {
55                              printf("%d step(s)\n",count);
56                             break;
57                         }
58                         if(s[r][c]=='N')
59                             r--;
60                         else if(s[r][c]=='S')
61                             r++;
62                         else if(s[r][c]=='W')
63                             c--;
64                         else if(s[r][c]=='E')
65                             c++;
66                     }
67                     break;
68                 }
69             }
70     }
71     return 0;
72 }

View Code

poj 2993 Emag eht htiw Em Pleh

http://poj.org/problem?id=2993

  1 #include<stdio.h>
  2 #include<string.h>
  3 int main()
  4 {
  5    int i,j,k;
  6     char s[17][33];
  7     memset(s,'.',sizeof(s));//棋盘初始化
  8     for(k=0; k<17; k+=2)
  9         for(j=0; j<33; j+=4)
 10             s[k][j]='+';
 11     for(k=1; k<17; k+=2)
 12         for(j=0; j<33; j+=4)
 13             s[k][j]='|';
 14     for(k=0; k<17; k+=2)
 15         for(j=1; j<33; j+=4)
 16             s[k][j]='-';
 17     for(k=0; k<17; k+=2)
 18         for(j=2; j<33; j+=4)
 19             s[k][j]='-';
 20     for(k=0; k<17; k+=2)
 21         for(j=3; j<33; j+=4)
 22             s[k][j]='-';
 23     for(k=1; k<17; k+=4)
 24         for(j=6; j<33; j+=8)
 25             s[k][j]=':';
 26     for(k=1; k<17; k+=4)
 27         for(j=5; j<33; j+=8)
 28             s[k][j]=':';
 29     for(k=1; k<17; k+=4)
 30         for(j=7; j<33; j+=8)
 31             s[k][j]=':';
 32     for(k=3; k<17; k+=4)
 33         for(j=2; j<33; j+=8)
 34             s[k][j]=':';
 35     for(k=3; k<17; k+=4)
 36         for(j=1; j<33; j+=8)
 37             s[k][j]=':';
 38     for(k=3; k<17; k+=4)
 39         for(j=3; j<33; j+=8)
 40             s[k][j]=':';
 41     char str1[20];
 42     char str2[100];
 43     scanf("%s %s",str1,str2);//输入棋子信息
 44     if(strcmp(str1,"White:") == 0)
 45     {
 46         for(i = 0; str2[i]; i++)
 47         {
 48             if(str2[i] == ',')
 49                 continue;
 50             else if(str2[i]>='A' && str2[i] <= 'Z')
 51             {
 52                 int y = str2[i+1]-'a';
 53                 int x = str2[i+2]-'0';
 54                 int row = (8-x)*2 + 1;
 55                 int col = y*4 + 2;
 56                 s[row][col] = str2[i];
 57                 i += 2;
 58             }
 59             else if(str2[i] >= 'a' && str2[i] <= 'z' && str2[i-1] == ',')
 60             {
 61                 int y = str2[i]-'a';
 62                 int x = str2[i+1]-'0';
 63                 int row = (8-x)*2 + 1;
 64                 int col = y*4 + 2;
 65                 s[row][col] = 'P';
 66                 i += 1;
 67             }
 68         }
 69     }
 70     scanf("%s %s",str1,str2);
 71     if(strcmp(str1,"Black:") == 0)
 72     {
 73         for(i = 0; str2[i]; i++)
 74         {
 75             if(str2[i] == ',')
 76                 continue;
 77             else if(str2[i]>='A' && str2[i] <= 'Z')
 78             {
 79                 int y = str2[i+1]-'a';
 80                 int x = str2[i+2]-'0';
 81                 int row = (8-x)*2 + 1;
 82                 int col = y*4 + 2;
 83                s[row][col] = str2[i]+32;
 84                 i += 2;
 85             }
 86             else if(str2[i] >= 'a' && str2[i] <= 'z' && str2[i-1] == ',')
 87             {
 88                 int y = str2[i]-'a';
 89                 int x = str2[i+1]-'0';
 90                 int row = (8-x)*2 + 1;
 91                 int col = y*4 + 2;
 92                 s[row][col] = 'p';
 93                 i += 1;
 94             }
 95         }
 96     }
 97     for(i=0; i<17; i++)
 98     {
 99         for(j=0; j<33; j++)
100             printf("%c",s[i][j]);
101         printf("\n");
102     }
103     return 0;
104 }

View Code

poj 2996 Help Me with the Game

http://poj.org/problem?id=2996

  1 #include<stdio.h>
  2 #include<string.h>
  3 #include<stdlib.h>
  4 struct node
  5 {char f,h,l;};
  6 int main()
  7 {
  8     int i,j;
  9     node w[100];
 10     node b[100];
 11     node p[100];
 12     char s[50][100];
 13     for(i=0; i<17; i++)
 14         scanf("%s%*c",s[i]);
 15     int count=0,cot=0;
 16     for(i=0; i<17; i++)
 17     {
 18         for(j=0; j<33; j++)
 19         {
 20             if(s[i][j]>='A'&&s[i][j]<='Z')
 21             {
 22                 w[count].f=s[i][j];
 23                 if(i==16)
 24                     w[count].h='1';
 25                 else
 26                     w[count].h='1'+(16-i)/2;
 27                 if(j==2)
 28                     w[count].l='a';
 29                 else
 30                     w[count].l='a'+(j-2)/4;
 31                 count++;
 32             }
 33             else if(s[i][j]>='a'&&s[i][j]<='z')
 34             {
 35                 b[cot].f=s[i][j];
 36                 if(i==16)
 37                     b[cot].h='1';
 38                 else
 39                     b[cot].h='1'+(16-i)/2;
 40                 if(j==2)
 41                     b[cot].l='a';
 42                 else
 43                     b[cot].l='a'+(j-2)/4;
 44                 cot++;
 45             }
 46         }
 47     }
 48     node k;
 49     for( i=0; i<count-1 ; i++ )
 50         for(j=i+1; j<count; j++)
 51         {
 52             if(w[i].f>w[j].f)
 53             {
 54                 k=w[i];
 55                 w[i]=w[j];
 56                 w[j]=k;
 57             }
 58             else if(w[i].f==w[j].f)
 59             {
 60                 if(w[i].h>w[j].h)
 61                 {
 62                     k=w[i];
 63                     w[i]=w[j];
 64                     w[j]=k;
 65                 }
 66                 else if(w[i].h==w[j].h)
 67                 {
 68                     if(w[i].l>w[j].l)
 69                     {
 70                         k=w[i];
 71                         w[i]=w[j];
 72                         w[j]=k;
 73                     }
 74                 }
 75             }
 76         }
 77     printf("White: ");
 78     for(i=0; i<count; i++)
 79         if(w[i].f=='K')
 80         {
 81             printf("%c%c%c,",w[i].f,w[i].l,w[i].h);
 82             break;
 83         }
 84     for(i=0; i<count; i++)
 85         if(w[i].f=='Q')
 86         {
 87             printf("%c%c%c,",w[i].f,w[i].l,w[i].h);
 88             break;
 89         }
 90     for(i=0; i<count; i++)
 91         if(w[i].f=='R')
 92             printf("%c%c%c,",w[i].f,w[i].l,w[i].h);
 93     for(i=0; i<count; i++)
 94         if(w[i].f=='B')
 95             printf("%c%c%c,",w[i].f,w[i].l,w[i].h);
 96     for(i=0; i<count; i++)
 97         if(w[i].f=='N')
 98             printf("%c%c%c,",w[i].f,w[i].l,w[i].h);
 99     int a;
100     a=0;
101     for(i=0; i<count; i++)
102       {
103           if(w[i].f=='P')
104             p[a++]=w[i];
105       }
106       for(i=0;i<a;i++)
107       {
108           if(i==a-1)
109             printf("%c%c\n",p[i].l,p[i].h);
110          else  printf("%c%c,",p[i].l,p[i].h);
111       }
112     printf("Black: ");
113     for(i=0; i<cot; i++)
114         if(b[i].f=='k')
115         {
116             printf("K%c%c,",b[i].l,b[i].h);
117             break;
118         }
119     for(i=0; i<cot; i++)
120         if(b[i].f=='q')
121         {
122             printf("Q%c%c,",b[i].l,b[i].h);
123             break;
124         }
125     for(i=0; i<cot; i++)
126         if(b[i].f=='r')
127             printf("R%c%c,",b[i].l,b[i].h);
128     for(i=0; i<cot; i++)
129         if(b[i].f=='b')
130             printf("B%c%c,",b[i].l,b[i].h);
131     for(i=0; i<cot; i++)
132         if(b[i].f=='n')
133             printf("N%c%c,",b[i].l,b[i].h);
134     a=0;
135     for(i=0; i<cot; i++)
136       {
137           if(b[i].f=='p')
138             p[a++]=b[i];
139       }
140       for(i=0;i<a;i++)
141       {
142           if(i==a-1)
143                     printf("%c%c\n",p[i].l,p[i].h);
144                 else  printf("%c%c,",p[i].l,p[i].h);
145       }
146     return 0;
147 }

View Code

模拟题不需要算法，只需严格按照题意将代码敲出来即可，过程一般比较麻烦，需要考虑的情况比较多。

posted on 2013-08-08 15:57 Fann0221 阅读(246) 评论(0) 编辑收藏举报