2022-07-09：总长度为n的数组中，所有长度为k的子序列里，有多少子序列的和为偶数？

2022-07-09：总长度为n的数组中，所有长度为k的子序列里，有多少子序列的和为偶数？

答案2022-07-09：

方法一：递归，要i还是不要i。 方法二：动态规划。需要两张dp表。

代码用rust编写。代码如下：

use rand::Rng;
fn main() {
    let nn: i32 = 20;
    let vv: i32 = 30;
    let test_time: i32 = 3000;
    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, n) + 1;
        let mut arr = random_array(n, vv);
        let ans1 = number1(&mut arr, k);
        let ans2 = number2(&mut arr, k);
        if ans1 != ans2 {
            println!("出错了!{}", i);
            println!("ans1 = {}", ans1);
            println!("ans2 = {}", ans2);
            break;
        }
    }
    println!("测试结束");
}

fn number1(arr: &mut Vec<i32>, k: i32) -> i32 {
    if arr.len() == 0 || k < 1 || k > arr.len() as i32 {
        return 0;
    }
    return process1(arr, 0, k, 0);
}

fn process1(arr: &mut Vec<i32>, index: i32, rest: i32, sum: i32) -> i32 {
    if index == arr.len() as i32 {
        return if rest == 0 && (sum & 1) == 0 { 1 } else { 0 };
    } else {
        return process1(arr, index + 1, rest, sum)
            + process1(arr, index + 1, rest - 1, sum + arr[index as usize]);
    }
}

fn number2(arr: &mut Vec<i32>, k: i32) -> i32 {
    if arr.len() == 0 || k < 1 || k > arr.len() as i32 {
        return 0;
    }
    let n = arr.len() as i32;
    // even[i][j] : 在前i个数的范围上(0...i-1)，一定选j个数，加起来是偶数的子序列个数
    // odd[i][j]  : 在前i个数的范围上(0...i-1)，一定选j个数，加起来是奇数的子序列个数
    let mut even: Vec<Vec<i32>> = vec![];
    let mut odd: Vec<Vec<i32>> = vec![];
    for i in 0..n + 1 {
        even.push(vec![]);
        odd.push(vec![]);
        for _ in 0..k + 1 {
            even[i as usize].push(0);
            odd[i as usize].push(0);
        }
    }
    for i in 0..=n {
        // even[0][0] = 1;
        // even[1][0] = 1;
        // even[2][0] = 1;
        // even[n][0] = 1;
        even[i as usize][0] = 1;
    }
    for i in 1..=n {
        for j in 1..=get_min(i, k) {
            even[i as usize][j as usize] = even[(i - 1) as usize][j as usize];
            odd[i as usize][j as usize] = odd[(i - 1) as usize][j as usize];
            even[i as usize][j as usize] += if (arr[(i - 1) as usize] & 1) == 0 {
                even[(i - 1) as usize][(j - 1) as usize]
            } else {
                odd[(i - 1) as usize][(j - 1) as usize]
            };
            odd[i as usize][j as usize] += if (arr[(i - 1) as usize] & 1) == 0 {
                odd[(i - 1) as usize][(j - 1) as usize]
            } else {
                even[(i - 1) as usize][(j - 1) as usize]
            };
        }
    }
    return even[n as usize][k as usize];
}

fn get_min<T: Clone + Copy + std::cmp::PartialOrd>(a: T, b: T) -> T {
    if a < b {
        a
    } else {
        b
    }
}

// 为了测试
fn random_array(n: i32, v: i32) -> Vec<i32> {
    let mut ans: Vec<i32> = vec![];
    for _i in 0..n {
        ans.push(rand::thread_rng().gen_range(0, v));
    }
    return ans;
}

执行结果如下：

左神java代码