Write code to complete DoublePennies()'s base case. Sample output for below program with inputs 1 and 10: Number of pennies after 10 days: 1024 Note: If the submitted code has an infinite loop, the system will stop running the code after a few seconds, and report "Program end never reached." The system doesn't print the test case that caused the reported message

Write code to complete DoublePennies()'s base case. Sample output for below program with inputs 1 and 10:

Number of pennies after 10 days: 1024

Note: If the submitted code has an infinite loop, the system will stop running the code after a few seconds, and report "Program end never reached." The system doesn't print the test case that caused the reported message

#include <stdio.h>

// Returns number of pennies if pennies are doubled numDays times
long long DoublePennies(long long numPennies, int numDays){
long long totalPennies;

/* Your solution goes here */

else {
totalPennies = DoublePennies((numPennies * 2), numDays - 1);
}

return totalPennies;
}

// Program computes pennies if you have 1 penny today,
// 2 pennies after one day, 4 after two days, and so on
int main(void) {
long long startingPennies;
int userDays;

scanf("%lld", &startingPennies);
scanf("%d", &userDays);
printf("Number of pennies after %d days: %lld\n", userDays, DoublePennies(startingPennies, userDays));

return 0;
}