spoj PUCMM025 solution method-2

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HII guys this is a simple problem...... U can do it as how we divide any number logic:: step 1: take any array for hash for counting how many times any digit is.... i.e int a[10] //for digits 0 to 9 and initilize it with 0 i.e intially count of every digit is zero(0) step 2: after this just check for every i from 0 to 9 if a[i]>0 then inside this just take s=0; for(j=0;j<len;j++) { s=s*10+(str[j]-'0'); if(s>=i) then s=s%i } now after this just check if s==0 then no stored in string is divisible by i th digit else not I HOPE YOU GOT THE LOGIC STILL IF U HAVE ANY PROBLEM U CAN COMMENT BELOW..... MY AC SOLUTION IS ........... #include<bits/stdc++.h> using namespace std; int main() { char str[260]; while(scanf("%s",str)!=EOF) { int a[10]={0}; int len=strlen(str),count=0,i,j; for(i=0;i<len;i++) { int index=str[i]-'0'; a[index]++; } if(a[1]>0) count+=a[1]; for(i=2;i<10;i++) { if(a[i]>0) { int s=0; // printf("dev"); for(j=0;j<len;j++) { s=s*10+(str[j]-'0'); if(s>=i) s%=i; } if(s==0) count+=a[i]; } } printf("%d\n",count); } return 0; }