2022-07-11：给定n位长的数字字符串和正数k，求该子符串能被k整除的子串个数。

use rand::Rng;
fn main() {
    let nn: i64 = 18;
    let kk: i64 = 20;
    let test_time: i32 = 3000;
    println!("测试开始");
    for i in 0..test_time {
        let str = random_number(rand::thread_rng().gen_range(0, nn) + 1);
        let k = rand::thread_rng().gen_range(0, kk) + 1;
        let ans1 = mod_ways1(&str, k);
        let ans2 = mod_ways2(&str, k);
        if ans1 != ans2 {
            println!("出错了!{}", i);
            println!("ans1 = {}", ans1);
            println!("ans2 = {}", ans2);
            break;
        }
    }
    println!("测试结束");
}

// 暴力方法
// 为了验证
fn mod_ways1(s: &str, k: i64) -> i64 {
    let n = s.len() as i64;
    let mut ans = 0;
    for i in 0..n {
        for j in i..n {
            if (&s[i as usize..(j + 1) as usize]).parse::<i64>().unwrap() % k == 0 {
                ans += 1;
            }
        }
    }
    return ans;
}

// 正式方法
// 时间复杂度O(N * k)
fn mod_ways2(s: &str, k: i64) -> i64 {
    let mut cur: Vec<i64> = vec![];
    for _ in 0..k {
        cur.push(0);
    }
    // 帮忙迁移
    let mut next: Vec<i64> = vec![];
    for _ in 0..k {
        next.push(0);
    }
    // 0...i 整体余几？
    let mut mod0: i64 = 0;
    // 答案：统计有多少子串的值%k == 0
    let mut ans = 0;
    for cha in s.bytes() {
        for i in 0..k {
            // i -> 10个
            // (i * 10) % k
            next[((i * 10) % k) as usize] += cur[i as usize];
            cur[i as usize] = 0;
        }
        let mut tmp = cur.clone();
        cur = next.clone();
        next = tmp.clone();
        mod0 = (mod0 * 10 + (cha - '0' as u8) as i64) % k;
        ans += if mod0 == 0 { 1 } else { 0 } + cur[mod0 as usize];
        cur[mod0 as usize] += 1;
    }
    return ans;
}

// 为了测试
fn random_number(n: i64) -> String {
    let mut ans: Vec<u8> = vec![];
    for _i in 0..n {
        ans.push('0' as u8 + rand::thread_rng().gen_range(0, 10));
    }
    return String::from_utf8(ans).unwrap();
}