原本数组中都是大于0、小于等于k的数字，是一个单调不减的数组，
其中可能有相等的数字，总体趋势是递增的。
但是其中有些位置的数被替换成了0，我们需要求出所有的把0替换的方案数量：
1）填充的每一个数可以大于等于前一个数，小于等于后一个数；
2）填充的每一个数不能大于k。

方法一：动态规划。
方法二：数学方法。用到组合，C(b-a+m,m)。

代码用rust编写。代码如下：

use rand::Rng;
fn main() {
    let nn: i64 = 20;
    let kk: i64 = 30;
    let test_time: i32 = 10000;
    println!("测试开始");
    for i in 0..test_time {
        let n = rand::thread_rng().gen_range(0, nn) + 1;
        let k = rand::thread_rng().gen_range(0, kk) + 1;
        let mut arr = random_array(n, k);
        let ans1 = ways1(&mut arr, k);
        let ans2 = ways2(&mut arr, k);
        if ans1 != ans2 {
            println!("出错了!{}", i);
            println!("ans1 = {}", ans1);
            println!("ans2 = {}", ans2);
            break;
        }
    }
    println!("测试结束");
}

// 动态规划
fn ways1(nums: &mut Vec<i64>, k: i64) -> i64 {
    let n = nums.len() as i64;
    // dp[i][j] : 一共i个格子，随意填，但是不能降序，j种数可以选
    let mut dp: Vec<Vec<i64>> = vec![];
    for i in 0..n + 1 {
        dp.push(vec![]);
        for _ in 0..k + 1 {
            dp[i as usize].push(0);
        }
    }
    for i in 1..=n {
        dp[i as usize][1] = 1;
    }
    for i in 1..=k {
        dp[1][i as usize] = i;
    }
    for i in 2..=n {
        for j in 2..=k {
            dp[i as usize][j as usize] =
                dp[(i - 1) as usize][j as usize] + dp[i as usize][(j - 1) as usize];
        }
    }
    let mut res = 1;
    let mut i: i64 = 0;
    let mut j: i64 = 0;
    while i < nums.len() as i64 {
        if nums[i as usize] == 0 {
            j = i + 1;
            while j < nums.len() as i64 && nums[j as usize] == 0 {
                j += 1;
            }
            let left_value = if i - 1 >= 0 {
                nums[(i - 1) as usize]
            } else {
                1
            };
            let right_value = if j < nums.len() as i64 {
                nums[j as usize]
            } else {
                k
            };
            res *= dp[(j - i) as usize][(right_value - left_value + 1) as usize];
            i = j;
        }
        i += 1;
    }
    return res;
}

// 数学方法
// a ~ b范围的数字随便选，可以选重复的数，一共选m个
// 选出有序序列的方案数：C ( m, b - a + m )
fn ways2(nums: &mut Vec<i64>, k: i64) -> i64 {
    let mut res = 1;
    let mut i: i64 = 0;
    let mut j: i64 = 0;
    while i < nums.len() as i64 {
        if nums[i as usize] == 0 {
            j = i + 1;
            while j < nums.len() as i64 && nums[j as usize] == 0 {
                j += 1;
            }
            let left_value = if i - 1 >= 0 {
                nums[(i - 1) as usize]
            } else {
                1
            };
            let right_value = if j < nums.len() as i64 {
                nums[j as usize]
            } else {
                k
            };
            let numbers = j - i;
            res *= c(right_value - left_value + numbers, numbers);
            i = j;
        }
        i += 1;
    }
    return res;
}

// 从一共a个数里，选b个数，方法数是多少
fn c(a: i64, b: i64) -> i64 {
    if a == b {
        return 1;
    }
    let mut x = 1;
    let mut y = 1;
    let mut i = b + 1;
    let mut j = 1;
    while i <= a {
        x *= i;
        y *= j;
        let gcd = gcd(x, y);
        x /= gcd;
        y /= gcd;
        i += 1;
        j += 1;
    }
    return x / y;
}

fn gcd(m: i64, n: i64) -> i64 {
    if n == 0 {
        m
    } else {
        gcd(n, m % n)
    }
}

// 为了测试
fn random_array(n: i64, k: i64) -> Vec<i64> {
    let mut ans: Vec<i64> = vec![];
    for _i in 0..n {
        ans.push(rand::thread_rng().gen_range(0, k) + 1);
    }
    ans.sort();
    for i in 0..n {
        ans[i as usize] = if rand::thread_rng().gen_range(0, 2) == 0 {
            0
        } else {
            ans[i as usize]
        };
    }
    return ans;
}

执行结果如下：