2023-01-04：有三个题库ABC，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个题目的难度为ai 题库B中第i个题目的难度为bi 题库C中第i个题目

Posted 2023-01-06 福大大架构师每日一题

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个题目的难度为ai
题库B中第i个题目的难度为bi
题库C中第i个题目的难度为ci
小美准备组合出一套试题，试题共有三道题，
第一题来自题库A，第二题来自题库B，第三题来自题库C
试题要求题目难度递增，且梯度不能过大
具体地说，第二题的难度必须大于第一题的难度，但不能大于第一题难度的两倍
第三题的难度必须大于第二题的难度，但不能大于第二题难度的两倍
小美想知道在满足上述要求下，有多少种不同的题目组合
（三道题目中只要存在一道题目不同，则两个题目组合就视为不同
输入描述 第一行一个正整数n, 表示每个题库的题目数量
第二行为n个正整数a1, a2,… an，其中ai表示题库A中第i个题目的难度值
第三行为n个正整数b1, b2,… bn，其中bi表示题库B中第i个题目的难度值
第四行为n个正整数c1, c2,… cn，其中ci表示题库C中第i个题目的难度值
1 <= n <= 20000, 1 <= ai, bi, ci <= 10^9。
来自美团。

答案2023-01-04：

双指针不回退+前缀和数组。
时间复杂度O(N * logN)。因为要排序。
空间复杂度O(N)。
用rust和solidity写代码。

代码用rust编写。代码如下：

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;

代码用solidity编写。代码如下：

// 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;