NBUT 1219 Time

标签：字符串   模拟   

• [1219] Time

• 时间限制: 1000 ms　内存限制: 131072 K
• 问题描述
• Digital clock use 4 digits to express time, each digit is described by 3*3 characters (including”|”,”_”and” “).now given the current time, please tell us how can it be expressed by the digital clock.
  
• 输入
• There are several test cases.
  Each case contains 4 integers in a line, separated by space.
  Proceed to the end of file.
• 输出
• For each test case, output the time expressed by the digital clock such as Sample Output.
• 样例输入

• 1 2 5 6
  2 3 4 2

• 样例输出

•     _  _  _ 
    | _||_ |_ 
    ||_  _||_|
   _  _     _ 
   _| _||_| _|
  |_  _|  ||_ 

• 提示

• The digits showed by the digital clock are as follows:
     _  _     _  _  _  _  _  _ 
   | _| _||_||_ |_   ||_||_|| |
   ||_  _|  | _||_|  ||_| _||_|

• 来源

• 辽宁省赛2010

  
  
  题目意思也很简单就是用字符串模拟时钟，注意每个数字占三格，最暴力的方法就是直接定义三个字符数组，按要求输出，还有一点就是0在最后一位，要特殊处理一下。
  
  

  #include <algorithm>
  #include <iostream>
  #include <cstring>
  #include <cstdlib>
  #include <cstdio>
  #include <string>
  #include <vector>
  #include <cmath>
  #include <ctime>
  #include <queue>
  #include <stack>
  #include <set>
  #include <map>
  using namespace std;
  typedef long long LL;
  const int maxn= 100000 + 10;
  
  char s1[35]="     _  _     _  _  _  _  _  _ ";
  char s2[35]="   | _| _||_||_ |_   ||_||_|| |";
  char s3[35]="   ||_  _|  | _||_|  ||_| _||_|";
  int main() {
      int a,b,c,d;
      while(~scanf("%d%d%d%d",&a,&b,&c,&d))
      {
          getchar();
          if(a==0)
              a=10;
          if(b==0)
              b=10;
          if(c==0)
              c=10;
          if(d==0)
              d=10;
          printf("%c%c%c%c%c%c%c%c%c%c%c%c\n",s1[a*3-2],s1[a*3-1],s1[a*3],s1[b*3-2],s1[b*3-1],s1[b*3],s1[c*3-2],s1[c*3-1],s1[c*3],s1[d*3-2],s1[d*3-1],s1[d*3]);
          printf("%c%c%c%c%c%c%c%c%c%c%c%c\n",s2[a*3-2],s2[a*3-1],s2[a*3],s2[b*3-2],s2[b*3-1],s2[b*3],s2[c*3-2],s2[c*3-1],s2[c*3],s2[d*3-2],s2[d*3-1],s2[d*3]);
          printf("%c%c%c%c%c%c%c%c%c%c%c%c\n",s3[a*3-2],s3[a*3-1],s3[a*3],s3[b*3-2],s3[b*3-1],s3[b*3],s3[c*3-2],s3[c*3-1],s3[c*3],s3[d*3-2],s3[d*3-1],s3[d*3]);
      }
      return 0;
  }
  

  
  这是最开始写的纯暴力方法，看上去好不协调啊…+_+…
  
  

  #include <algorithm>
  #include <iostream>
  #include <cstring>
  #include <cstdlib>
  #include <cstdio>
  #include <string>
  #include <vector>
  #include <cmath>
  #include <ctime>
  #include <queue>
  #include <stack>
  #include <set>
  #include <map>
  using namespace std;
  typedef long long LL;
  const int maxn= 100000 + 10;
  
  char  s[3][35]= {"     _  _     _  _  _  _  _  _ ",
                   "   | _| _||_||_ |_   ||_||_|| |",
                   "   ||_  _|  | _||_|  ||_| _||_|"
                  };
  int a[4];
  int main() {
      while(~scanf("%d",&a[0])) {
          for(int i=1; i<4; i++)
              scanf("%d",&a[i]);
          for(int i=0; i<4; i++)
              if(!a[i])
                  a[i]+=10;
          for(int i=0; i<3; i++) {
              for(int j=0; j<4; j++)
              {
                  int t=a[j];
                  printf("%c%c%c",s[i][t*3-2],s[i][t*3-1],s[i][t*3]);
              }
              puts("");
          }
      }
      return 0;
  }
  

  
  O(∩_∩)O哈哈~这样看上去清爽多了吧。
  
  
  还有一种比较屌的方法，是照着基神的代码敲的（真不知道基神脑袋怎么长的，就是跟常人思维不一样……好吧…%>_<%…其实不可否认的还是自己太愚笨…Orz），虽然本人目前没看懂，但还是贴上来把，求大神赐教^_^
  
  

  #include <algorithm>
  #include <iostream>
  #include <cstring>
  #include <cstdlib>
  #include <cstdio>
  #include <string>
  #include <vector>
  #include <cmath>
  #include <ctime>
  #include <queue>
  #include <stack>
  #include <set>
  #include <map>
  using namespace std;
  typedef long long LL;
  const int maxn= 100000 + 10;
  
  char  s[50][50];
  int data[]={119,36,93,109,46,107,123,37,127,111};
  
  int slove(int p,int x)
  {
      x=data[x];
      if(x>>0&1)s[0][p+1]='_';
      if(x>>1&1)s[1][p+0]='|';
      if(x>>2&1)s[1][p+2]='|';
      if(x>>3&1)s[1][p+1]='_';
      if(x>>4&1)s[2][p+0]='|';
      if(x>>5&1)s[2][p+2]='|';
      if(x>>6&1)s[2][p+1]='_';
  }
  
  int main() {
      int a[5];
      while(~scanf("%d",&a[0])) {
          for(int i=1; i<4; i++)
              scanf("%d",&a[i]);
          memset(s,' ',sizeof(s));
          for(int i=0;i<3;i++)
              s[i][12]=0;
          for(int i=0;i<4;i++)
              slove(i*3,a[i]);
          for(int i=0;i<3;i++)
              printf("%s\n",s[i]);
      }
      return 0;
  }
  

  
  

NBUT 1219 Time

标签：字符串   模拟   

原文地址：http://blog.csdn.net/zhang_xueping/article/details/47751293