用数组输入n个数的方法（2022-12-22给定一个数字n）

2022-12-22：给定一个数字n，代表数组的长度，

给定一个数字m，代表数组每个位置都可以在1~m之间选择数字，

所有长度为n的数组中，最长递增子序列长度为3的数组，叫做达标数组。

返回达标数组的数量。

1 <= n <= 500，

1 <= m <= 10，

500 * 10 * 10 * 10，

结果对998244353取模，

实现的时候没有取模的逻辑，因为非重点。

来自微众银行。

答案2022-12-22：

参考最长递增子序列。

代码用rust编写。代码如下：

use std::iter::repeat; fn main() { println!("功能测试开始"); for n in 4..=8 { for m in 1..=5 { let ans1 = number1(n, m); let ans2 = number2(n, m); if ans1 != ans2 { println!("{}", ans1); println!("{}", ans2); println!("出错了!"); } } } println!("功能测试结束"); } // 暴力方法 // 为了验证 fn number1(n: i32, m: i32) -> i32 { let mut a: Vec<i32> = repeat(0).take(n as usize).collect(); return process1(0, n, m, &mut a); } fn process1(i: i32, n: i32, m: i32, path: &mut Vec<i32>) -> i32 { if i == n { return if length_of_lis(path) == 3 { 1 } else { 0 }; } else { let mut ans = 0; for cur in 1..=m { path[i as usize] = cur; ans = process1(i 1, n, m, path); } return ans; } } fn length_of_lis(arr: &mut Vec<i32>) -> i32 { if arr.len() == 0 { return 0; } let mut ends: Vec<i32> = repeat(0).take(arr.len()).collect(); ends[0] = arr[0]; let mut right = 0; let mut max = 1; for i in 1..arr.len() as i32 { let mut l = 0; let mut r = right; while l <= r { let mut m = (l r) / 2; if arr[i as usize] > ends[m as usize] { l = m 1; } else { r = m - 1; } } right = get_max(right, l); ends[l as usize] = arr[i as usize]; max = get_max(max, l 1); } return max; } fn get_max<T: Clone Copy std::cmp::PartialOrd>(a: T, b: T) -> T { if a > b { a } else { b } } // i : 当前来到的下标 // f、s、t : ends数组中放置的数字！ // ? == 0，没放！ // n : 一共的长度！ // m : 每一位，都可以在1~m中随意选择数字 // 返回值：i..... 有几个合法的数组！ fn zuo(i: i32, f: i32, s: i32, t: i32, n: i32, m: i32) -> i32 { if i == n { return if f != 0 && s != 0 && t != 0 { 1 } else { 0 }; } // i < n let mut ans = 0; for cur in 1..=m { if f == 0 || f >= cur { ans = zuo(i 1, cur, s, t, n, m); } else if s == 0 || s >= cur { ans = zuo(i 1, f, cur, t, n, m); } else if t == 0 || t >= cur { ans = zuo(i 1, f, s, cur, n, m); } } return ans; } // 正式方法 // 需要看最长递增子序列！ // 尤其是理解ends数组的意义！ fn number2(n: i32, m: i32) -> i32 { //repeat(vec![]).take((m 1) as usize).collect(); let mut dp: Vec<Vec<Vec<Vec<i32>>>> = repeat( repeat( repeat(repeat(-1).take((m 1) as usize).collect()) .take((m 1) as usize) .collect(), ) .take((m 1) as usize) .collect(), ) .take(n as usize) .collect(); return process2(0, 0, 0, 0, n, m, &mut dp); } fn process2( i: i32, f: i32, s: i32, t: i32, n: i32, m: i32, dp: &mut Vec<Vec<Vec<Vec<i32>>>>, ) -> i32 { if i == n { return if f != 0 && s != 0 && t != 0 { 1 } else { 0 }; } if dp[i as usize][f as usize][s as usize][t as usize] != -1 { return dp[i as usize][f as usize][s as usize][t as usize]; } let mut ans = 0; for cur in 1..=m { if f == 0 || cur <= f { ans = process2(i 1, cur, s, t, n, m, dp); } else if s == 0 || cur <= s { ans = process2(i 1, f, cur, t, n, m, dp); } else if t == 0 || cur <= t { ans = process2(i 1, f, s, cur, n, m, dp); } } dp[i as usize][f as usize][s as usize][t as usize] = ans; return ans; }