## Tuesday, 3 April 2018

### hackerrank solution: Project Euler #34: Digit factorials

solution : python2

# Enter your code here. Read input from STDIN. Print output to STDOUT
a = [0]*10
a[0]=1

for i in range(1,10):
a[i] = int(a[i-1] * i)

n = int(input())
ans = 0

for i in range(10,n+1):
temp = int(i)
sum = 0
while temp > 0 :
sum += a[int(temp%10)]
temp//=10

sum = int(sum)
if sum % i == 0 :
ans += i

print (int(ans))

### Project Euler #11: Largest product in a grid

HACKERRANK SOLUTION: c++

#include<iostream>
using namespace std;
int main(){
long long a[20][20],max=0,p;
for(int i=0;i<20;i++){
for(int j=0;j<20;j++){
cin>>a[i][j];
}
}
for(int i=0;i<20;i++){
for(int j=0;j<=16;j++){
p = a[i][j]*a[i][j+1]*a[i][j+2]*a[i][j+3];
if(p> max) max=p;

p=a[j][i]*a[j+1][i]*a[j+2][i]*a[j+3][i];
if(p> max) max=p;
}
}

for(int i=0;i<17;i++){
for(int j=0;j<17;j++){
p= a[i][j]*a[i+1][j+1]*a[i+2][j+2]*a[i+3][j+3];
if(p> max) max=p;
}
}
for(int i=3;i<20;i++){
for(int j=0;j<17;j++){

p=a[i][j]*a[i-1][j+1]*a[i-2][j+2]*a[i-3][j+3];
if(p> max) max=p;
}
}

cout<<max;
return 0;

}

### Project Euler #38: Pandigital multiples; Hackerrank

Take the number  and multiply it by each of , and
By concatenating each product we get the  to  pandigital, . We will call  the concatenated product of  and
The same can be achieved by starting with  and multiplying by , and , giving the pandigital, , which is the concatenated product of  and . Let's call 9 as the Multiplier
The similar process can be shown for  to  pandigital also.  when multiplied by  gives  which is  pandigital.
You are given  and  where  =  or , find the multipliers for that given  below  and print them in ascending order.
Input Format
Input contains two integer  and .
Constraints

Output Format
Print the answer corresponding to the test case.
Sample Input
100 8

Sample Output
18
78
solution: c++
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define f(a,b,c)                for(int a=b;a<c;a++)
#define s(x)                    scanf("%d",&x);
#define sl(x)                   scanf("%lld",&x);
#define p(x)                    printf("%d\n",x);
#define p2(x,y)                 printf("%d %d\n",x,y);
#define pl(x)                   printf("%lld\n",x);
#define pl2(x,y)                 printf("%lld %lld\n",x,y);
#define p1d(a,n)                for(int ix=0;ix<n;ix++) printf("%d ",a[ix]); printf("\n");
#define p2d(a,n,m)              for(int ix=0;ix<n;ix++){ for(int jx=0;jx<m;jx++) printf("%d ",a[ix][jx]); printf("\n");}
void input(){
#ifdef Megamind
#define DEBUG
#define TRACE
#define gc getchar()
freopen("inp.txt","r",stdin);
//freopen("out1.txt","w",stdout);
#else
#define gc getchar_unlocked()
#endif
}
#ifdef TRACE
#define trace(x)                       cerr<<__FUNCTION__<<":"<<__LINE__<<": "#x" = "<<x<<endl;
#define trace2(x,y)                    cerr<<__FUNCTION__<<":"<<__LINE__<<": "#x" = "<<x<<" | "#y" = "<<y<<endl;
#define trace3(x,y,z)                  cerr<<__FUNCTION__<<":"<<__LINE__<<": "#x" = "<<x<<" | "#y" = "<<y<<" | "#z" = "<<z<<endl;
#else
#define trace(x)
#define trace2(x,y)
#define trace3(x,y,z)
#endif
template <class T>
n1=0;
char c=gc;
while(c<'0' || c>'9') c=gc;
while(c>='0' && c<='9'){
n1=(n1<<3)+(n1<<1)+c-'0';
c=gc;
}
}
inline ll power(ll a, ll b, ll m) {
ll r = 1;
while(b) {
if(b & 1) r = r * a % m;
a = (a * a)% m;
b >>= 1;
}
return r;
}
inline ll power(ll a, ll b) {
ll r = 1;
while(b) {
if(b & 1) r = r * a ;
a = a * a;
b >>= 1;
}
return r;
}

/*........................................................END OF TEMPLATES.......................................................................*/
int digit[10];

bool check(int num , int k){

bool ok = 1;
int temp ,mod,counter=0;
memset(digit,0,sizeof(digit));

f(i,1,10){
if(counter >= k) break;
temp = num*i;

while(temp){

mod = temp%10;

if(!mod || mod>k || digit[mod])  return !ok;
digit[mod]=1;

counter++;
temp/=10;
}
}

return (counter == k);

}

int main(){
input();
int n,k;

s(n);
s(k);
f(i,2,n)
if(check(i,k))
p(i);

#ifdef Megamind
cout << "Time elapsed: "<< 1.0 * clock() / CLOCKS_PER_SEC * 1000 << " ms." << endl;
#endif
}