leeyupeng

leeyupeng@126.com

0%

发表于 更新于 分类于 ， ， 阅读次数： Valine：

栈

栈是一种后进先出的数据结构

s.push压栈，s.pop弹出栈顶,s.top获取栈顶元素

栈实现

Stack.hpp
Stack.cpp

函数调用栈

递归与栈

递归函数的实现是基于函数栈实现的

递归可以使用栈来实现，通过栈保存当前环境

单调栈

阅读全文 »

发表于 更新于 分类于 ， 阅读次数： Valine：

函数模板

类模板

模板的非类型参数

bitset_1.h

模板的实例化

隐式实例化

显示实例化

模板的特化

模板的继承

阅读全文 »

发表于 更新于 分类于 ， 阅读次数： Valine：

发表于 更新于 分类于 ， 阅读次数： Valine：

union

在C/C++程序的编写中，当多个基本数据类型或复合数据结构要占用同一片内存时，我们要使用联合体；当多种类型，多个对象，多个事物只取其一时（我们姑且通俗地称其为“n 选1”），我们也可以使用联合体来发挥其长处。

vector4 用来表示向量和颜色 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
struct  vector4
{
	union
	{
		struct {
			float r;
			float g;
			float b;
			float a;
		};
		struct
		{
			float x;
			float y;
			float z;
			float w;
		};
	};
};
1
2
3
4
5
6
7
8
9
10
11
12
void test() {
	vector4 v4;
	v4.x = 1.0f;
	v4.y = 2.0f;
	cout << v4.x << " " << v4.y << endl;
	cout << v4.r << " " << v4.g << endl;

	v4.r = 2.0f;
	v4.g = 1.0f;
	cout << v4.x << " " << v4.y << endl;
	cout << v4.r << " " << v4.g << endl;
}
阅读全文 »

发表于 更新于 分类于 ， 阅读次数： Valine：

1363F Rotating Substrings

问题描述

You are given two strings s and t, each of length n and consisting of lowercase Latin alphabets. You want to make s equal to t.

You can perform the following operation on s any number of times to achieve it —

Choose any substring of s and rotate it clockwise once, that is, if the selected substring is s[l,l+1…r], then it becomes s[r,l,l+1…r−1]. All the remaining characters of s stay in their position.
For example, on rotating the substring [2,4] , string “abcde” becomes “adbce”.

A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Find the minimum number of operations required to convert s to t, or determine that it’s impossible.

Input
The first line of the input contains a single integer t (1≤t≤2000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1≤n≤2000) — the length of the strings.

阅读全文 »