Which XXX and YYY correctly output the smallest values? Vector user Vals contains integers (which may be positive or negative).

Question

1 year ago

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Which XXX and YYY correctly output the smallest values? Vector user Vals contains integers (which may be positive or negative).

Choices are in the form XXX/YYY. // Determine smallest (min) value int minval; XXX for (i = 0; i < uservals.size(); ++i) { if (YYY) { minval - userVals.at(i); cout << "Min: " << minval << endl; minval - uservals.at(); /uservals.at(i) < minval minval = 0; /userval > minval minval - uservals.at(); /uservals.at(i) > minval minval - 0; /userval < minval

Computers and Technology

1 answer:

liubo4ka [24] · Answer 1 · 2024-03-27T10:39:08+03:00

Answer:

The answer is "minVal - userVals.at(0); /userVals.at(i) < minVal "

Explanation:

In the question, it uses minVal instead of XXX to hold the very first arra(userVal) element, and rather than YYY you choose a conditional statement to check which integer is lower than minVal to index of loop increment. It changes the value of the minVal if the condition is valid. It's completely different if we talk about another situation.