最近学习到的奇技淫巧：Lambda表达式，将函数包括递归函数改为Lambda表达式写法，可节省大量时间，在大量调用下可能节省近一半时间。

说明

该语法过于复杂，见https://en.cppreference.com/w/cpp/language/lambda，本文仅写在算法竞赛下的应用。

该语法在OIWiki中有所提及，但是十分抽象，而这里将给出的简单易懂的用法，可能不太全面，在算法竞赛中已经够用了。

有关该语法是否可用问题：关于NOI系列活动中编程语言使用限制的补充说明，这表明NOI系列比赛中（包括noip，csp）已经开始使用C++14标准，而该表达式在C++11中就已经支持

具体用法：

无自身递归调用

auto 函数名 = [&](参数) -> 函数类型 { 内容 }；

给定 x 、y, 求x + y

一般写法

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

int sum(int x, int y)
{
	return x + y;
}

int main()
{
	int x, y;
	cin >> x >> y;
	cout << sum(x, y);
}

Lambda表达式写法

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

int main()
{
	auto sum = [&](int x, int y) -> int 
	{
		return x + y;
	};//注意这里的分号
	int x, y;
	cin >> x >> y;
	cout << sum(x, y);
}

有自身函数调用

注意：如果函数内存在对自生的调用，按上述写法是无法编译的，我们需要这样写：

int main()
{
	auto 函数名 = [&](参数, auto&& self) -> 函数类型
	{ 
		//内容 
		//对自身调用时：
		self(参数， self);
	}；
	//主函数内调用：
	函数名（参数， 函数名）;//举例 ： int d = gcd(x, y, gcd);
};

gcd函数

Lambda表达式写法

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

int main()
{
	auto gcd = [&](int x, int y, auto&& self) -> int 
	{
		if(y == 0) return x;
		else return self(y, x % y, self);
	};//注意这里的分号
	int x, y;
	cin >> x >> y;
	cout << gcd(x, y, gcd);
}

线段树

题目luogu线段树1

一般代码

// Problem: P3372 【模板】线段树 1
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3372
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define ll long long
#define MAXn 100010
#define ls(x) tr[x].ls
#define rs(x) tr[x].rs
#define sum(x) tr[x].sum
#define add(x) tr[x].add
#define l(x) tr[x].l
#define r(x) tr[x].r
#define sz(x) tr[x].sz
#define mid(x) tr[x].mid
using namespace std;

int n, m;
int a[MAXn];

struct SegmentTree
{
	ll ls, rs, l, r, sum, add, sz, mid;
} tr[MAXn << 2];

ll read()
{
	ll num = 0, w = 1;
	char ch = getchar();
	while(ch > '9' || ch < '0') { if(ch == '-') w = -1; ch = getchar(); }
	while(ch <= '9' && ch >= '0') { num = (num << 3) + (num << 1) + (ch - '0');	ch = getchar(); }
	return num * w;
}

void build(int ind, int L, int R)
{
	l(ind) = L;
	r(ind) = R;
	sz(ind) = R - L + 1;
	mid(ind) = (L + R) >> 1;
	if(L == R)
	{
		sum(ind) = a[L];
		return;
	}
	ls(ind) = ind << 1;
	rs(ind) = ind << 1 | 1;
	int mid = (L + R) >> 1;
	build(ls(ind), L, mid);
	build(rs(ind), mid + 1, R);
	sum(ind) = sum(ls(ind)) + sum(rs(ind));
}

void pushdown(int x)
{
	if(add(x))
	{
		sum(ls(x)) += add(x) * sz(ls(x));
		sum(rs(x)) += add(x) * sz(rs(x));
		add(ls(x)) += add(x);
		add(rs(x)) += add(x);
		add(x) = 0;
	}
}

void update(int x, int L, int R, ll k)
{
	if(L <= l(x) && R >= r(x))
	{
		sum(x) += sz(x) * k;
		add(x) += k;
		return;
	}
	pushdown(x);
	if(L <= mid(x)) update(ls(x), L, R, k);
	if(R >= mid(x) + 1) update(rs(x), L, R, k);
	sum(x) = sum(ls(x)) + sum(rs(x));
}

ll ask(int x, int L, int R)
{
	if(L <= l(x) && R >= r(x))
	{
		return sum(x);
	}
	pushdown(x);
	ll res = 0;
	if(L <= mid(x)) res += ask(ls(x), L, R);
	if(R >= mid(x) + 1) res += ask(rs(x), L, R);
	return res;
}

int main()
{
	n = read(), m = read();
	for(int i = 1; i <= n; i++)
		a[i] = read();
	build(1, 1, n);
	while(m--)
	{
		ll flag, x, y;
		flag = read(), x = read(), y = read();
		if(flag == 1)
		{
			ll k;
			k = read();
			update(1, x, y, k);
		}
		else 
		{
			printf("%lld\n", ask(1, x, y));
		}
	}
	return 0;
}

Lambda代码

// Problem: P3372 【模板】线段树 1
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3372
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define ll long long
#define MAXn 100010
#define ls(x) tr[x].ls
#define rs(x) tr[x].rs
#define sum(x) tr[x].sum
#define add(x) tr[x].add
#define l(x) tr[x].l
#define r(x) tr[x].r
#define sz(x) tr[x].sz
#define mid(x) tr[x].mid
using namespace std;

int n, m;
int a[MAXn];

struct SegmentTree
{
	ll ls, rs, l, r, sum, add, sz, mid;
} tr[MAXn << 2];

int main()
{
	
	auto read = [&]() -> ll
	{
		ll num = 0, w = 1;
		char ch = getchar();
		while(ch > '9' || ch < '0') { if(ch == '-') w = -1; ch = getchar(); }
		while(ch <= '9' && ch >= '0') { num = (num << 3) + (num << 1) + (ch - '0');	ch = getchar(); }
		return num * w;
	};
	auto build = [&](int ind, int L, int R, auto&& self) -> void
	{
		l(ind) = L;
		r(ind) = R;
		sz(ind) = R - L + 1;
		mid(ind) = (L + R) >> 1;
		if(L == R)
		{
			sum(ind) = a[L];
			return;
		}
		ls(ind) = ind << 1;
		rs(ind) = ind << 1 | 1;
		int mid = (L + R) >> 1;
		self(ls(ind), L, mid, self);
		self(rs(ind), mid + 1, R, self);
		sum(ind) = sum(ls(ind)) + sum(rs(ind));
	};
	auto pushdown = [&](int x) -> void
	{
		if(add(x))
		{
			sum(ls(x)) += add(x) * sz(ls(x));
			sum(rs(x)) += add(x) * sz(rs(x));
			add(ls(x)) += add(x);
			add(rs(x)) += add(x);
			add(x) = 0;
		}
	};
	auto update = [&](int x, int L, int R, ll k, auto&& self) -> void 
	{
		if(L <= l(x) && R >= r(x))
		{
			sum(x) += sz(x) * k;
			add(x) += k;
			return;
		}
		pushdown(x);
		if(L <= mid(x)) self(ls(x), L, R, k, self);
		if(R >= mid(x) + 1) self(rs(x), L, R, k, self);
		sum(x) = sum(ls(x)) + sum(rs(x));
	};
	auto ask = [&](int x, int L, int R, auto&& self) -> ll
	{
		if(L <= l(x) && R >= r(x))
		{
			return sum(x);
		}
		pushdown(x);
		ll res = 0;
		if(L <= mid(x)) res += self(ls(x), L, R, self);
		if(R >= mid(x) + 1) res += self(rs(x), L, R, self);
		return res;
	};

	n = read(), m = read();
	for(int i = 1; i <= n; i++)
		a[i] = read();
	build(1, 1, n, build);
	while(m--)
	{
		ll flag = read(), x = read(), y = read();
		if(flag == 1)
		{
			ll k;
			k = read();
			update(1, x, y, k, update);
		}
		else 
		{
			printf("%lld\n", ask(1, x, y, ask));
		}
	}
	return 0;
}

时间比较

• 一般做法

• Lambda表达式