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;
}
,
