Codu And Sum Love
Problem Description
```
Scanner sc = new Scanner(System.in);
long sum = 0;
int N = sc.nextInt();
for (int i = 0; i < N; i++) {
final long x = sc.nextLong(); // read input
String str = Long.toString((long) Math.pow(1 << 1, x));
str = str.length() > 2 ? str.substring(str.length() - 2) : str;
sum += Integer.parseInt(str);
}
System.out.println(sum%100);
```
Given N number of x’'s, perform logic equivalent of the above Java code and print the output
Constraints
1<=N<=10^7 0<=x<=10^18
Input Format
First line contains an integer N
Second line will contain N numbers delimited by space
Output
Number that is the output of the given code by taking inputs as specified above
Explanation
Example 1
Input
4 8 6 7 4
Output
64
Example 2
Input
3
1 2 3
Output
14
Program
#include <stdio.h>
long power(long k)
{
long i,j=1;
for(i=1;i<=k;i++)
j=j*2;
return j;
}
long length(long k)
{
long ct=0,n1;
while(k)
{
ct++;
k=k/10;
}
return ct;
}
long reduce(long p1,long r1)
{
long ct=0,k,cnt,rev=0,a[1000],h=0,i,sum1=0;
while(p1)
{
k=p1%10;
a[h]=k;
h++;
p1=p1/10;
}
for(i=1;i>=0;i--)
sum1=sum1*10+a[i];
return sum1;
}
int main() {
long x,n,sum=0,r[70],p,c,f=0,i;
scanf("%ld",&n);
for(i=0;i<n;i++)
{
scanf("%ld",&x);
p=power(x);
c=length(p);
if(c>2)
{
r[f++]=reduce(p,c);
}
else
r[f++]=p;
}
for(i=0;i<n;i++)
sum=sum+r[i];
printf("%ld",(sum%100));
return 0;
}
Output:
4 8 6 7 4
64
You can also try running it on online IDE: https://ide.geeksforgeeks.org/G3gXokhTbW
Your doubts and feedback are welcomed! you can comment it below Cheers!
Related Link: Super Ascii
