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Code Golf: Collatz
The Collatz conjecture states that, for any positive integer n, it will eventually reach 1 by repeatedly applying the following procedure:
- If n is even, divide it by 2.
- If n is odd, multiply by 3 and then add 1.
The number of steps needed for n to reach 1 is called its stopping time. For example, the stopping time of 10 is six:
10 → 5 → 16 → 8 → 4 → 2 → 1
Print the stopping times of all the numbers from 1 to 1,000 inclusive, each on their own line.
Obfuscated C
What does this obfuscated C code do? Make a guess as to its logic/why it works.
int main(int b,char**i){long long n=B,a=I^n,r=(a/b&a)>>4,y=atoi(*++i),_=(((a^n/b)*(y>>T)|y>>S)&r)|(a^r);printf("%.8s\n",(char*)&_);}
To compile
clang -include stdio.h -include stdlib.h -Wall -Weverything -pedantic -DB=6945503773712347754LL -DI=5859838231191962459LL -DT=0 -DS=7 -o prog prog.c
The compile-time flags of -DB, -DI, -DT, -DS
are passed to the program.
Hint: Look at the bitwise operations and split it out by line! Look at the variable _
.
More details here: burton IOCC
Figure it out
What does this code do? Explain why this program works.
#include <stdio.h>
#include <stdlib.h>
#define SIZEOF(arr) (sizeof(arr)/sizeof(arr[0]))
#define PrintInt(expr) printf("%s:%d\n",#expr,(expr))
int main()
{
/* The powers of 10 */
int pot[] = {
0001,
0010,
0100,
1000
};
int i;
for(i=0;i<SIZEOF(pot);i++)
PrintInt(pot[i]);
return 0;
}
Code Golf: Pernicious Numbers
A pernicious number is a positive number where the sum of its binary expansion is a prime number.
For example, 5 is a pernicious number since 5 = 1012 and 1 + 1 = 2, which is prime.
Print all the pernicious numbers from 0 to 50 inclusive, each on their own line. There are 35 such numbers.
Code Golf: Missing Digit
Given 9 digits from 0 to 9, write the shortest code you can in Python or C to find the missing digit.
Example:
Input : 756432098
Output: 1
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