A whole team is competing. Use a for loop over the vectors to find the most productive employee, then print their 15% bonus.
You start with:
let names = vec!["John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"];
let productivities = vec![83, 133, 97, 150, 120];
let salaries = vec![135000, 335000, 210000, 250000, 180000];
Build:
Alice Chen Bonus: $37500
Rules:
- Index
iin all three vectors refers to the same employee. - The winner is whichever employee has the highest productivity. If two employees share the highest productivity, the one listed first wins.
- The winner's bonus is 15% of their salary, rounded down to a whole number.
- Print the winner's name on the first line, then
"Bonus: $<amount>"on the second.
Start with
let mut bestIndex = 0 and loop over the other indices with for i in 1..names.len(). Inside the loop, update the best index only when productivities[i] > productivities[bestIndex]: a strict comparison keeps the first employee on a tie.
Dividing two whole numbers (
i32) drops the fraction, so salaries[bestIndex] * 15 / 100 is already rounded down: let bonus = salaries[bestIndex] * 15 / 100;
{} in the format string is replaced by the value: println!("Bonus: ${}", bonus);
let names = vec!["John Doe", "Mary Jane", "Bob Smith", "Alice Chen", "Carlos Diaz"];
let productivities = vec![83, 133, 97, 150, 120];
let salaries = vec![135000, 335000, 210000, 250000, 180000];
let mut bestIndex = 0;
for i in 1..names.len() {
if productivities[i] > productivities[bestIndex] {
bestIndex = i;
}
}
let bonus = salaries[bestIndex] * 15 / 100;
println!("{}", names[bestIndex]);
println!("Bonus: ${}", bonus);
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