递增三元组 多种做法

 

第一种做法：前缀和  O( n )

第二种做法：排序+二分  O(nlgn)

前缀和

空间换时间

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;

int a[N], b[N], c[N];
int cnta[N], cntc[N];
//cnta数组含义: cnta[i] a中i出现的次数
int main()
{
	int n;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]++;
	
	for (int i = 1;i <= n;i++)
		cnta[a[i]]++, cntc[c[i]]++;

	for (int i = 1;i <= N;i++)
	{//这里ABC数组里最大值时10的五次方
		cnta[i] += cnta[i - 1];
		cntc[i] += cntc[i - 1];
	}
	long long sum = 0;
	for (int i = 1;i <= n;i++)
		sum += (long long)cnta[b[i]-1] * (cntc[N-1]-cntc[b[i]]);
//sum += (long long)cnta[b[i]-1] * (n-cntc[b[i]]);//也可以这样写
	cout << sum;
	system("pause");
}

二分+排序 

这里的二分还得处理特殊情况 （  if (a[l] >= x) l = 0; 根据结果的正确性返回一个值 ）

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
int a[N], b[N], c[N];
int n;
int find(int x)
{
	int l = 1;int r = n;
	if (a[l] >= x) l = 0;

	while (l < r)
	{

		int mid = l + r + 1 >> 1;
		if (a[mid] < x) l = mid;
		else r = mid - 1;
	}
	return l;//返回下标
}

int find2(int x)
{
	int l = 1;int r = n;
	if (c[r] <= x) l = n+1;

	while (l < r)
	{

		int mid = l + r  >> 1;
		if (c[mid] > x) r = mid;
		else l = mid + 1;
	}
	return l;//返回下标
}


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];
	sort(a+1, a + n+1);
	sort(c+1, c + n+1);


	long long sum = 0;
	for (int i = 1;i <= n;i++)
	{
		sum = sum + (long long)find(b[i]) * (n+1-find2(b[i]));

	}
	cout << sum;
	system("pause");
}

 二分统一形式👇

#include<bits/stdc++.h>
using namespace std;
const int N = 100010;
int a[N], b[N], c[N];
int n;
int find(int x,int a[])
{
	int l = 1;int r = n;
	if (a[l] >= x) l = 0;

	while (l < r)
	{

		int mid = l + r + 1 >> 1;
		if (a[mid] < x) l = mid;
		else r = mid - 1;
	}
	return l;//返回下标
}


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];
	sort(a + 1, a + n + 1);
	sort(c + 1, c + n + 1);


	long long sum = 0;
	for (int i = 1;i <= n;i++)
	{
		sum = sum + (long long)find(b[i],a) * (n  - find(b[i]+1,c));

	}
	cout << sum;
	system("pause");
}

直接调用upper_bound 和lower_bound

#include<bits/stdc++.h>
using namespace std;

typedef long long LL;
const int N = 1e5 + 10;
int num[3][N];

int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < 3; ++i)
        for (int j = 1; j <= n; ++j)
            scanf("%d", &num[i][j]);
    for (int i = 0; i < 3; ++i)
        sort(num[i] + 1, num[i] + n + 1);

    LL ans = 0;
    for (int i = 1; i <= n; ++i) {
        int key = num[1][i];
        //A中二分查找第一个小于key的数的下标
        int pos1 = lower_bound(num[0] + 1, num[0] + n + 1, key) - num[0] - 1;
        //C中二分查找第一个大于key的数的下标
        int pos2 = upper_bound(num[2] + 1, num[2] + n + 1, key) - num[2];
        if (pos1 >= 1 && pos2 <= n) {
            ans += (LL)pos1 * (n - pos2 + 1);
        }
    }
    cout << ans << endl;
    system("pause");
    return 0;
}

二分模板

 模板还是要记住的，毕竟上面的那个函数只能处理大于等于 不等价于这里的check

如何记忆理解：对于如何判断用r=mid or l=mid

我们不妨先从左到右遍历一下区间 并代入l=mid 与check 的条件 判断是否会筛去我们想要的结果值。若筛去了，则从右到左遍历区间  并代入r=mid 与check的条件 判断是否会筛去（若前者筛去了这里就不会筛去了）

字符串转数字 

外加一个“2019”字符串转数字 2019的方法 (stoi 也可以

int x=0;
for( i = 0 ; i < s.size() ; i++ )
    x=x*10 + str[i]-' 0 ';