题目：
1.主要涉及字符串处理，把一些凌乱的句子转换为正常的句子。
2.字母之间只能够是空格或者逗号+空格或者句号+空格隔开。
3.一行里面可能有多个句子，这些句子不用换行，但是需要用句号+空格隔开，并且第一个字母需要大写。
4.注意最后结尾只能是句号，如果还包含有空格，则会出现Presentation Error。
4.主要是上面的情况，下面是一个测试用例及其正确的答案，这个测试用例正确，就能AC。
/*
测试用例：
my Name is Little Hi. ami .i hate hiho ,codr. EEENNN.
His name IS Little ho , We are friends.
my Name is Little Hi.His name IS Little ho , We are friends.
my Name is Little Hi. His name IS Little ho , We are friends.
His name IS Little ho ,We are friends. weoifj ,jo.
Jo, f, j l, o,o, l.
正确答案：
My name is little hi. Ami. I hate hiho, codr. Eeennn.
His name is little ho, we are friends.
My name is little hi. His name is little ho, we are friends.
My name is little hi. His name is little ho, we are friends.
His name is little ho, we are friends. Weoifj, jo.
Jo, f, j l, o, o, l.
*/
时间限制:10000ms
单点时限:1000ms
内存限制:256MB
描述
To prepare for the English exam Little Ho collected many digital reading materials. Unfortunately the materials are messed up by a malware.
It is known that the original text contains only English letters (a-zA-Z), spaces, commas, periods and newlines, conforming to the following format:
1. Each sentence contains at least one word, begins with a letter and ends with a period.
2. In a sentence the only capitalized letter is the first letter.
3. In a sentence the words are separated by a single space or a comma and a space.
4. The sentences are separated by a single space or a single newline.
It is also known the malware changes the text in the following ways:
1. Changing the cases of letters.
2. Adding spaces between words and punctuations.
Given the messed text, can you help Little Ho restore the original text?
输入
A string containing no more than 8192 English letters (a-zA-Z), spaces, commas, periods and newlines which is the messed text.
输出
The original text.
- 样例输入
-
my Name is Little Hi.
His name IS Little ho , We are friends.
- 样例输出
-
My name is little hi.
His name is little ho, we are friends.
AC代码：
/*
题目：
1.主要涉及字符串处理。
2.字母之间只能够是空格或者逗号+空格或者句号+空格隔开。
3.一行里面可能有多个句子，这些句子不用换行，但是需要用句号+空格隔开，并且第一个字母需要大写。
*/
/*
测试用例：
my Name is Little Hi. ami .i hate hiho ,codr. EEENNN.
His name IS Little ho , We are friends.
my Name is Little Hi.His name IS Little ho , We are friends.
my Name is Little Hi. His name IS Little ho , We are friends.
His name IS Little ho ,We are friends. weoifj ,jo.
Jo, f, j l, o,o, l.
正确答案：
My name is little hi. Ami. I hate hiho, codr. Eeennn.
His name is little ho, we are friends.
My name is little hi. His name is little ho, we are friends.
My name is little hi. His name is little ho, we are friends.
His name is little ho, we are friends. Weoifj, jo.
Jo, f, j l, o, o, l.
*/
#include<string>
#include <iomanip>
#include<fstream>
#include<set>
#include<queue>
#include<map>
//#include<unordered_set>
//#include<unordered_map>
//#include <sstream>
//#include "func.h"
//#include <list>
#include<stdio.h>
#include<iostream>
#include<string>
#include<memory.h>
#include<limits.h>
//#include<stack>
#include<vector>
#include <algorithm>
using namespace std;
char lower(char&c)
{
if (c <= 'Z'&&c >= 'A') return c - 'A' + 'a';
else return c;
}
char upper(char&c)
{
if (c <= 'z'&&c >= 'a') return c - 'a' + 'A';
else return c;
}
/*
函数名 ：main
函数功能：主函数
*/
int main(void)
{
string str;
while (getline(cin, str))
{
string newStr="";
bool blank = false;
bool comma = false;
bool period = false;
for (int i = 0; i < str.size(); i++)
{
//如果是句子的开头，且str[i]元素不为空
if (newStr == ""&&str[i]!=' ')
{
newStr += upper(str[i]);
}
//如果不是句子的开头
else if (newStr != "")
{
//如果是句号
if (str[i] == '.')
{//标记之前记录的是句号
period = true;
newStr += str[i];
}
else if (str[i] == ',')
{
comma = true;
newStr += str[i];
}
//标记空白
else if (str[i]==' ')
blank = true;
else
{
//如果之前是句号，则这个是新的句子，要加上空白，并且字母要大写
if (period)
{
newStr += ' ';
newStr += upper(str[i]);
period = false;
comma = false;
blank = false;
}
//如果之前是逗号或者是空白，则要加上空白，字母要小写
else if (comma || blank)
{
newStr += ' ';
newStr += lower(str[i]);
period = false;
comma = false;
blank = false;
}
//其他情况，即只是字母，全部小写
else
newStr += lower(str[i]);
}
}
}
cout << newStr << endl;
}
return 0;
}
，一共是2^k个数。这2^k个数中，二进制中最高位为1的是第k+1位，且所有2^k个数的最高位均为1。
，设n的最高位1是在第k+1位(k>=2)，
n’记为：
当n%4=2时，即n=n’+2，所以
