PROJECT EULER SOLUTION # 43

Solution to problem number 43 of Project Euler.

Question # 43

The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.
Let d₁ be the 1^st digit, d₂ be the 2^nd digit, and so on. In this way, we note the following:

• d₂d₃d₄=406 is divisible by 2
• d₃d₄d₅=063 is divisible by 3
• d₄d₅d₆=635 is divisible by 5
• d₅d₆d₇=357 is divisible by 7
• d₆d₇d₈=572 is divisible by 11
• d₇d₈d₉=728 is divisible by 13
• d₈d₉d₁₀=289 is divisible by 17

Find the sum of all 0 to 9 pandigital numbers with this property.

Solution # 43

/********************************************************************/

#include<stdio.h>

int isPandigital(long long int);

int isSatisfying(long long int);

int main()

{

      int a,b,c,d,e,f,g,h,i,j;

      long long int num,sum=0;

      for(a=1;a<10;a++)

       for(num=a,b=0;b<10;b++)

        if(isPandigital(num=a*10+b))

         for(c=0;c<10;c++)

          if(isPandigital(num=a*100+b*10+c))

           for(d=0;d<10;d++)

            if(isPandigital(num=a*1000+b*100+c*10+d))

             for(e=0;e<10;e++)

              if(isPandigital(num=a*10000+b*1000+c*100+10*d+e))

               for(f=0;f<10;f++)

                if(isPandigital(num=a*100000+10000*b+1000*c+100*d+10*e+f))

               for( g=0;g<10;g++)

                if(isPandigital(num=a*1000000+b*100000+c*10000+d*1000+e*100+f*10+g))

                 for(h=0;h<10;h++)

                  if(isPandigital(num=a*10000000+b*1000000+c*100000+d*10000+e*1000+f*100+g*10+h))

                   for(i=0;i<10;i++)

                  if(isPandigital(num=a*100000000ll+b*10000000ll+c*1000000ll+d*100000ll+e*10000ll+f*1000ll+g*100ll+h*10ll+i))

                   for(j=0;j<10;j++)

                    if(isPandigital(num=a*1000000000ll+b*100000000ll+c*10000000ll+d*1000000ll+e*100000ll+f*10000ll+g*1000ll+h*100ll+i*10ll+j))

                     if(isSatisfying(num))

                     {

                        printf("%lld\n",num);

                        sum+=num;

                     }

      printf("sum = %lld\n",sum);

      return 0;

}

int isSatisfying(long long int n)

{

      if((n%1000000000/1000000)%2!=0)

            return 0;

      if((n%100000000/100000)%3!=0)

            return 0;

      if((n%10000000/10000)%5!=0)

            return 0;

      if((n%1000000/1000)%7!=0)

            return 0;

      if((n%100000/100)%11!=0)

            return 0;

      if((n%10000/10)%13!=0)

            return 0;

      if((n%1000)%17!=0)

            return 0;

      return 1;

}

int isPandigital(long long int n)

{

      int arr[10]={0},i;

      while(n)

      {

            arr[n%10]++;

            n/=10;

      }

      for(i=0;i<10;i++)

            if(arr[i]>=2)

                  return 0;

      return 1;

}

/********************************************************************/