蓝桥杯每日一题2023.10.28

题目描述

递增三元组 - 蓝桥云课 (lanqiao.cn)

题目分析

60分解法：

直接暴力循环每一个数进行比较

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
typedef long long ll;
ll n, a[N], b[N], c[N], ans;
int main()
{
	cin >> n;
	for(int i = 1; i <= n; i ++)cin >> a[i];
	for(int i = 1; i <= n; i ++)cin >> b[i];
	for(int i = 1; i <= n; i ++)cin >> c[i];
	for(int i = 1; i <= n; i ++)
	{
		for(int j = 1; j <= n; j ++)
		{
			for(int k = 1; k <= n; k ++)
			{
				if(a[i] < b[j] && b[j] < c[k])ans ++;
			}
		}
	}
	cout << ans;
	return 0;
}

满分解法：

由于ABC的值是完全独立的所以可以使用乘法原理

由B作为一个判断点，看有多少个A符合要求，再看有多少个C符合要求，最后的答案则为两部分相乘的结果

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
ll n, a[N], b[N], c[N], s[N], cnt[N];
ll sa[N];//sa[i]表示在a[]中有多少个数小于b[i] 
ll sc[N];//sc[i]表示在c[]中有多少个数大于b[i]
int main()
{
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> n;
	for(int i = 1; i <= n; i ++)cin >> a[i], a[i] ++;
	for(int i = 1; i <= n; i ++)cin >> b[i], b[i] ++;
	for(int i = 1; i <= n; i ++)cin >> c[i], c[i] ++;
	//求sa[] 
	for(int i = 1; i <= n; i ++)cnt[a[i]] ++;
	for(int i = 1; i <= N; i ++)s[i] = s[i - 1] + cnt[i];
	for(int i = 1; i <= n; i ++)sa[i] = s[b[i] - 1];
	//求sc[] 
	memset(cnt, 0, sizeof cnt);
	memset(s, 0, sizeof s);
	for(int i = 1; i <= n; i ++)cnt[c[i]] ++;
	for(int i = 1; i <= N; i ++)s[i] = s[i - 1] + cnt[i];
	for(int i = 1; i <= n; i ++)sc[i] = s[N] - s[b[i]];
	//枚举每个b[i]
	ll ans = 0;
	for(int i = 1; i <= n; i ++)ans += sa[i] * sc[i];
	cout << ans;
	return 0;
}