20240913 finished.
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@ -234,3 +234,4 @@ mod p2181_merge_nodes_in_between_zeros;
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mod p2552_count_increasing_quadruplets;
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mod p2555_maximize_win_from_two_segments;
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mod p2576_find_the_maximum_number_of_marked_indices;
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mod p2398_maximum_number_of_robots_within_budget;
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66
src/problem/p2398_maximum_number_of_robots_within_budget.rs
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66
src/problem/p2398_maximum_number_of_robots_within_budget.rs
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@ -0,0 +1,66 @@
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/**
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* [2398] Maximum Number of Robots Within Budget
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*/
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pub struct Solution {}
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// submission codes start here
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impl Solution {
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pub fn maximum_robots(charge_times: Vec<i32>, running_costs: Vec<i32>, budget: i64) -> i32 {
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use std::collections::VecDeque;
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let n = charge_times.len();
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let mut result = 0;
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let mut running_costs_sum = 0i64;
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let mut queue = VecDeque::with_capacity(n);
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// 最左侧的机器人
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let mut left = 0;
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for i in 0..n {
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running_costs_sum += running_costs[i] as i64;
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// 维持单调队列
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while let Some(&pos) = queue.back() {
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if charge_times[pos] <= charge_times[i] {
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queue.pop_back();
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} else {
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break;
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}
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}
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queue.push_back(i);
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// 判断当前left开始的机器人是否会导致超过预算
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while left <= i && (i - left + 1) as i64 * running_costs_sum + charge_times[*queue.front().unwrap()] as i64 > budget {
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if let Some(&front) = queue.front() {
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if front == left {
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queue.pop_front();
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}
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}
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running_costs_sum -= running_costs[left] as i64;
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left += 1;
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}
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if i >= left {
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result = result.max(i - left + 1);
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}
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}
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result as i32
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}
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}
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// submission codes end
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#[cfg(test)]
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mod tests {
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use super::*;
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#[test]
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fn test_2398() {
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assert_eq!(3, Solution::maximum_robots(vec![3, 6, 1, 3, 4], vec![2, 1, 3, 4, 5], 25));
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assert_eq!(0, Solution::maximum_robots(vec![11, 12, 19], vec![10, 8, 7], 19));
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}
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}
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