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Sunday 18 January 2015

spoj PUCMM025 solution 

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hii guys this is logic for spoj Divisor Digits problem.............
1.every no is divisble by 1
2.Any no is divisible by 2 if its last digit is divisible by 2
3.Any no is divisible by 3 if its sum of digit is divisible by 3
4.Any no is divisible by 4 if number formed by its last two digits is divisible by 4
5.Any no is divisible by 5 if its last digit is divisible by 5  i.e its last digit either 0 or 5
6.Any no is divisible by 6 if the no is divisible by 2 and 3
8.Any no is divisible by 8 if number formed by last three digit is divisible by 8
9.Any no is divisible by 9 if sum of digits is divisible by 9

The only issue is how to check for 7...
so there is a very good concept for 7........
I am declareing an array
int a[]={1,3,2,-1,-3,-2};
letsay i want to check for str[]="73089089";
its len=8;
start from Rightmost digit upto leftmost digit
and just find out the sum as follows.......
int sum=0,k=0;
for(int i=len-1;i>=0;i--)
{
sum+=(str[i]-' 0')*(a[k]);
k=(k+1)%6;
}
after this you have to check
if(sum%7==0)
then number is divisible by 7
else
not


I HOPE YOU GOT THE LOGIC............







my AC solution is.........

#include <cstdio>
#include <cstring>

int main()
{
 int len,count,sum,num,num2,i;
 int array[] = {1,3,2,-1,-3,-2};
 int yes;
 char str[245],chr;
 while(scanf("%s",str) != EOF)
 {
  len = strlen(str);
  for(i = 0,count = 0,sum = 0,num = 0,num2 = 0,yes = 0; i < len; i++)
  {
   chr = str[i];
   if(chr == '0')
   {
    continue;
   }
   else if(chr == '1')
   {
    count++;
   }
   else if(chr == '2')
   {
    if((str[len-1]-48)%2 == 0)
     count++;
   }
   else if(chr == '3')
   {
    if(sum == 0)
    {
     for(int j = 0; j < len; j++)
     {
      sum += (str[j] - 48);
     }
    }
    if(sum % 3 == 0)
     count ++;
   }
   else if(chr == '4')
   {
    if(num == 0)
    {
     num = 10*(str[len-2] - 48) + (str[len-1] - 48);
    }
    if(num%4 == 0)
     count++;
   }
   else if(chr == '5')
   {
    if((str[len-1]-48)%5 == 0)
     count++;
   }
   else if(chr == '6')
   {
    if(sum == 0)
    {
     for(int j = 0; j < len; j++)
     {
      sum += (str[j] - 48);
     }
    }
    if(sum%3 == 0 && (str[len-1]-48)%2 == 0)
     count++;
   }
   else if(chr == '7')
   {
    if(yes == 1) count++;
    else if(yes == -1) continue;
    else
    {
     int numk,k = 0,nsum = 0;
     for(int j = len-1; j >=0 ; j--)
     {
      numk = str[j] - 48;
      nsum += array[k]*numk;
      k = (k+1)%6;
     }
     if(nsum% 7 == 0)
     {
      count++;
      yes = 1;
     }
     else yes = -1;
    }
   }
   else if(chr == '8')
   {
    if(num2 == 0)
    {
     num2 = 100*(str[len-3]-48) + 10*(str[len-2]-48) + str[len-1]-48;
    }
    if(num2%8 == 0)
     count++;
   }
   else if(chr == '9')
   {
    if(sum == 0)
    {
     for(int j = 0; j < len; j++)
     {
      sum += (str[j] - 48);
     }
    }
    if(sum % 9 == 0)
     count++;
   }
  }
  printf("%d\n",count);
 }
}
at

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