2023-02-12：给定正数 N，表示用户数量，用户编号从 0~N-1， 给定正数 M，表示实验数量，实验编号从 0~M-1， 给定长度为 N 的二维数组 A， A[i] = { a, b, c }表示，用户 i 报

use rand::Rng;use std::collections::HashMap;use std::collections::HashSet;use std::iter::repeat;fn main() {    let N = 100;    let M = 20;    let Q = 50;    let testTime = 5000;    println!("功能测试开始");    for i in 0..testTime {        let n = rand::thread_rng().gen_range(0, N) + 1;        let m = rand::thread_rng().gen_range(0, M) + 1;        let mut A = randomMatrix(n, rand::thread_rng().gen_range(0, m) + 1, m);        let q = rand::thread_rng().gen_range(0, Q) + 1;        let mut B = randomMatrix(q, rand::thread_rng().gen_range(0, m) + 1, m);        let ans1 = record1(n, m, q, &mut A, &mut B);        let ans2 = record2(n, m, q, &mut A, &mut B);        let mut pass = true;        for j in 0..q {            if ans1[j as usize] != ans2[j as usize] {                pass = false;                break;            }        }        if !pass {            println!("出错了!");            break;        }    }    println!("功能测试结束");}
// 暴力方法// 为了验证fn record1(n: i32, m: i32, q: i32, A: &mut Vec<Vec<i32>>, B: &mut Vec<Vec<i32>>) -> Vec<i32> {    let mut expUsersMap: HashMap<i32, HashSet<i32>> = HashMap::new();    for i in 0..m {        expUsersMap.insert(i, HashSet::new());    }    for i in 0..n {        for exp in A[i as usize].iter() {            let mut a = expUsersMap.get_mut(exp).unwrap();            a.insert(i);        }    }    let mut ans: Vec<i32> = repeat(0).take(q as usize).collect();    let mut help: HashSet<i32> = HashSet::new();    for i in 0..q {        help.clear();        for exp in B[i as usize].iter() {            for user in expUsersMap.get(exp).unwrap().iter() {                help.insert(*user);            }        }        ans[i as usize] = help.len() as i32;    }    return ans;}
// 正式方法fn record2(n: i32, m: i32, q: i32, A: &mut Vec<Vec<i32>>, B: &mut Vec<Vec<i32>>) -> Vec<i32> {    // n 一共有多少人    // 任何一个实验，需要几个整数，能表示所有人谁出现谁没出现？    let parts = (n + 31) / 32;    // m    0 ~ m -1    // [i]  [.........]    let mut bitMap: Vec<Vec<i32>> = repeat(repeat(0).take(parts as usize).collect())        .take(m as usize)        .collect();    for i in 0..n {        // i 人的编号 : a b c        for exp in A[i as usize].iter() {            bitMap[*exp as usize][(i / 32) as usize] |= 1 << (i % 32);        }    }    let mut ans: Vec<i32> = repeat(0).take(q as usize).collect();    for i in 0..q {        // i号查询 ： a、c、e，一共有多少去重的人        // a[0] | c[0] | e[0] -> 几个1        // a[1] | c[1] | e[1] -> 几个1        let mut all = 0;        for j in 0..parts {            let mut status = 0;            for exp in B[i as usize].iter() {                status |= bitMap[*exp as usize][j as usize];            }            all += countOnes(status);        }        ans[i as usize] = all;    }    return ans;}
// 大厂刷题班，32节，leetcode专题 : https://leetcode.com/problems/number-of-1-bits/fn countOnes(mut n: i32) -> i32 {    let mut n2: u32;    if n < 0 {        n2 = (u32::MAX as i64 + n as i64 + 1) as u32;    } else {        n2 = n as u32;    }    let mut n = n2;    n = (n & 0x55555555) + ((n >> 1) & 0x55555555);    n = (n & 0x33333333) + ((n >> 2) & 0x33333333);    n = (n & 0x0f0f0f0f) + ((n >> 4) & 0x0f0f0f0f);    n = (n & 0x00ff00ff) + ((n >> 8) & 0x00ff00ff);    n = (n & 0x0000ffff) + ((n >> 16) & 0x0000ffff);    return n as i32;}
// 为了测试fn randomMatrix(n: i32, m: i32, v: i32) -> Vec<Vec<i32>> {    let mut ans: Vec<Vec<i32>> = repeat(repeat(0).take(m as usize).collect())        .take(n as usize)        .collect();    for i in 0..n {        for j in 0..m {            ans[i as usize][j as usize] = rand::thread_rng().gen_range(0, v);        }    }    return ans;}