Final answer:
The code will output '1 1' after execution because the value of 'funny' is set to 1 by the modulo operation and then later confirmed by the conditional statements which assign both 'funny' and 'serious' to 1.
Step-by-step explanation:
After the execution of the given code, the modulo operation is performed first: serious % 2, which results in serious being 15 and 15 % 2 equals 1. Therefore, funny is assigned the value of 1 since the modulo operation of an odd number by 2 always results in 1. Then, the code enters the if statement block. Since funny is equal to 1, the first condition if (funny != 1) does not hold true, so the code does not enter that block, and both funny and serious retain their current values. It then moves to the else if condition which checks if (funny == 2). This condition also evaluates to false because funny is equal to 1 and not 2. Finally, it moves to the else block, which adjusts the values of both funny and serious to 1. The output will thus display funny and serious as the numbers 1 and 1 respectively. Therefore, when cout << funny << " " << serious << endl; runs, it prints 1 1.