Codechef Code Crunch 2013 RD-CODE

RD-CODE Problem code: MRIU12

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Problem description.

Code Crunch Heads use RD-CODE for sending messages. It consists of only dots and dashes in sequence.
Below are the RD-CODE for the digits 0 to 9:

• 0 -----
• 1 .----
• 2 ..---
• 3 ...--
• 4 ....-
• 5 .....
• 6 -....
• 7 --...
• 8 ---..
• 9 ----.
  
  
  
  
  Using this code, help Code Crunch Members to find out the RD-CODE of their message.
  Users will enter 10 digit phone no. Convert these numbers to RD-CODE and also find no of dots used in the coded data.

  Input

  First line T, will have a single integer having number of test cases followed by T lines of input.
  
  
  
  
  Second line will have a 10-digit number N where N is a positive integer < 9999999999.

  Output

  First line will display the RD-Code for each digit together.
  
  
  Second line will display the total number of dots in the RD-CODE.

  Example

  Input:
  2
  9716733117
  9716335949
  .
  
  Output:
  ----.--….-----….--……--…--.----.------…
  23
  ----.--….-----…….--…--…..----…..-----.
  26

  #pragma comment(linker, "/STACK:1024000000,1024000000")
  #include <map>
  #include <queue>
  #include <vector>
  #include <string>
  #include <cstdio>
  #include <cstring>
  #include <iostream>
  #include <algorithm>
  using namespace std;
  #define maxn 10005
  #define mod 1000000007
  #define ll long long
  #define INF 0x7fffffff
  int n, m;
  char s[maxn];
  int main(){
      int cas = 1;
      int t;
      scanf("%d", &t);
      while (t--){
          scanf("%s", s);
          n = 0;
          char str[151] = "-----.----..---...--....-.....-....--...---..----.";
          for (int i = 0; i < strlen(s); i++){
              if (s[i]<='5')n += s[i] - '0';
              if (s[i] > '5')n += 5 - (s[i] - '0') % 5;
              for (int j = 0; j < 5; j++)printf("%c", str[(s[i] - '0') * 5 + j]);
          }
          printf("\n%d\n", n);
      }
      /*while (~scanf("%s", s)){
          
      }*/
      return 0;
  }