2023-01-04：有三个题库A、B、C，每个题库均有n道题目，且题目都是从1到n进行编号每个题目都有一个难度值题库A中第i个

use rand::Rng;
use std::iter::repeat;
fn main() {
    let mut a = vec![71, 29, 13, 74, 90, 8, 30, 25];
    let mut b = vec![72, 1, 50, 98, 86, 86, 52, 57];
    let mut c = vec![11, 24, 61, 70, 11, 90, 13, 30];
    let ans = ways2(&mut a, &mut b, &mut c);
    println!("ans = {}", ans);
    let nn: i32 = 100;
    let vv: i32 = 100;
    let test_time: i32 = 5000;
    println!("测试开始");
    for i in 0..test_time {
        let n = rand::thread_rng().gen_range(0, nn) + 1;
        let mut a = random_array(n, vv);
        let mut b = random_array(n, vv);
        let mut c = random_array(n, vv);
        let ans1 = ways1(&mut a, &mut b, &mut c);
        let ans2 = ways2(&mut a, &mut b, &mut c);
        if ans1 != ans2 {
            println!("出错了!{}", i);
            println!("ans1 = {}", ans1);
            println!("ans2 = {}", ans2);
            break;
        }
    }
    println!("测试结束");
}

// 暴力方法
// 时间复杂度O(N^3)
// 为了验证
fn ways1(a: &mut Vec<i32>, b: &mut Vec<i32>, c: &mut Vec<i32>) -> i32 {
    let n = a.len() as i32;
    a.sort();
    b.sort();
    c.sort();
    let mut ans = 0;
    for i in 0..n {
        let mut j = 0;
        while j < n && b[j as usize] <= a[i as usize] * 2 {
            if b[j as usize] > a[i as usize] {
                let mut k = 0;
                while k < n && c[k as usize] <= b[j as usize] * 2 {
                    if c[k as usize] > b[j as usize] {
                        ans += 1;
                    }
                    k += 1;
                }
            }
            j += 1;
        }
    }
    return ans;
}

// 正式方法
// 时间复杂度O(N * logN)
fn ways2(a: &mut Vec<i32>, b: &mut Vec<i32>, c: &mut Vec<i32>) -> i32 {
    let n = a.len() as i32;
    a.sort();
    b.sort();
    c.sort();
    // B里面的记录
    let mut help: Vec<i32> = repeat(0).take(n as usize).collect();
    let mut i = 0;
    let mut l = -1;
    let mut r = 0;
    while i < n {
        while l + 1 < n && c[(l + 1) as usize] <= b[i as usize] {
            l += 1;
        }
        while r < n && c[r as usize] <= b[i as usize] * 2 {
            r += 1;
        }
        help[i as usize] = get_max(r - l - 1, 0);
        i += 1;
    }
    for i in 1..n {
        help[i as usize] += help[(i - 1) as usize];
    }
    let mut ans = 0;
    let mut i = 0;
    let mut l = -1;
    let mut r = 0;
    while i < n {
        while l + 1 < n && b[(l + 1) as usize] <= a[i as usize] {
            l += 1;
        }
        while r < n && b[r as usize] <= a[i as usize] * 2 {
            r += 1;
        }
        if r - l - 1 > 0 {
            ans += sum(&mut help, l + 1, r - 1);
        }
        i += 1;
    }
    return ans;
}

fn sum(help: &mut Vec<i32>, l: i32, r: i32) -> i32 {
    return if l == 0 {
        help[r as usize]
    } else {
        help[r as usize] - help[(l - 1) as usize]
    };
}

fn get_max<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 arr: Vec<i32> = vec![];
    for _i in 0..n {
        arr.push(rand::thread_rng().gen_range(0, v));
    }
    return arr;
}

// SPDX-License-Identifier: MIT
pragma solidity ^0.8.17;

contract Hello{
    function main() public pure returns (int32){
        int32[] memory a = new int32[](8);
    a[0] = 71;
    a[1] = 29;
    a[2] = 13;
    a[3] = 74;
    a[4] = 90;
    a[5] = 8;
    a[6] = 30;
    a[7] = 25;
    int32[]  memory b = new int32[](8);
    b[0] = 72;
    b[1] = 1;
    b[2] = 50;
    b[3] = 98;
    b[4] = 86;
    b[5] = 86;
    b[6] = 52;
    b[7] = 57;
    int32[]  memory c = new int32[](8);
    c[0] = 11;
    c[1] = 24;
    c[2] = 61;
    c[3] = 70;
    c[4] = 11;
    c[5] = 90;
    c[6] = 13;
    c[7] = 30;
    int32 ans = ways2(a,b,c);
        return ans;
    }

  // 正式方法
  // 时间复杂度O(N * logN)
  function ways2(int32[] memory a, int32[] memory b, int32[] memory c) public pure returns (int32){
    int32 n = int32(int(a.length));
    sort(a);
    sort(b);
    sort(c);
    // B里面的记录
    int32[] memory help  = new int32[](uint(uint32(n)));
    int32 l = -1;
    int32 r = 0;
    for (int32 i = 0; i < n; i++) {
      while (l + 1 < n && c[uint(uint32(l + 1))] <= b[uint(uint32(i))]) {
        l++;
      }
      while (r < n && c[uint(uint32(r))] <= b[uint(uint32(i))] * 2) {
        r++;
      }
      help[uint(uint32(i))] = getMax(r - l - 1, 0);
    }
    for (int32 i = 1; i < n; i++) {
      help[uint(uint32(i))] += help[uint(uint32(i - 1))];
    }
    int32 ans = 0;
    l = -1;
    r = 0;
    for (int32 i = 0; i < n; i++) {
      while (l + 1 < n && b[uint(uint32(l + 1))] <= a[uint(uint32(i))]) {
        l++;
      }
      while (r < n && b[uint(uint32(r))] <= a[uint(uint32(i))] * 2) {
        r++;
      }
      if (r - l - 1 > 0) {
        ans += sum(help, l + 1, r - 1);
      }
    }
    return ans;
  }

  function sum(int32[] memory help, int32 l, int32 r) public pure returns (int32){
    return l == 0 ? help[uint(uint32(r))] : help[uint(uint32(r))] - help[uint(uint32(l - 1))];
  }

    function getMax(int32 a,int32 b) public pure returns (int32){
        if(a>b){
            return a;
        }else{
            return b;
        }
    }

  function sort(int32[] memory arr)public pure{
    int32 temp = 0;
    for(int32 i = 1; i <= int32(int(arr.length)) - 1; i++) {
      for(int32 j = i; j > 0; j--) {
        if(arr[uint(uint32(j - 1))] > arr[uint(uint32(j))]) {
          temp = arr[uint(uint32(j))];
          arr[uint(uint32(j))] = arr[uint(uint32(j - 1))];
          arr[uint(uint32(j - 1))] = temp; 
        }else {
          //不满足条件结束循环即可
          break;
        }
      }
    }
  }
}