Uva problem solution 11743 - Credit Check

problem link

Discuss: This is just easy applied problem.You can take input using %1d specifier or using char string.then easily calculate the value and check last digit using remainder.

try yourself before see the code


#include<bits/stdc++.h>
using namespace std;
int main()
{
  int a,b,c,d,n,i,j,sum,oddsum,totalsum,t,k,r;
  cin >>t;
  for(k=0;k<t;k++)
  {
  totalsum=0;
  for(i=0;i<4;i++)
  {
  sum=0;
    scanf("%1d%1d%1d%1d",&a,&b,&c,&d);
     a*=2;
     c*=2;
  while(a>0)
  {
    r=a%10;
    a/=10;
    sum+=r;

  }
  while(c>0)
  {
    r=c%10;
    c/=10;
    sum+=r;

  }
  sum+=b;
  sum+=d;
  totalsum+=sum;

  }
  r=totalsum%10;
  if(r==0)
  cout <<"Valid"<<endl;
  else
  cout <<"Invalid"<<endl;

}

}