HDU 4054 Hexadecimal View【模拟】【字符串处理】

Hexadecimal View

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2707    Accepted Submission(s): 1096

Problem Description

Hexadecimal is very important and useful for computer programmers. You are requested to provide a hexadecimal view for given data. The hexadecimal view is made up of one or more rows. Every row except the last one represents 16 characters. Each row consists of three columns separated by a space: 

* addr: the 4-digit hexadecimal beginning address of this row. 
* dump: the hexadecimal representation of this row, separating every two characters by a whitespace. If there are less than 16 characters in the last row, pad it with spaces. 
* text: the ASCII translation of this row, with uppercase characters converted to lowercase and lowercase characters converted to uppercase. 
Use lowercase for the letter digits. See sample for more details. 

 

Input

There are multiple test cases. Each line is a test case. The line is made up of no less than 1 and no more than 4096 printable characters including spaces. 

 

Output

For each test case, output its hexadecimal view. Do not output any extra spaces after the last character of text. 

 

Sample Input

Hex Dump #include <cstdio> printf("Hello, World!\n"); main = do getLine >>= print . sum . map read . words

 

Sample Output

0000: 4865 7820 4475 6d70 hEX dUMP 0000: 2369 6e63 6c75 6465 203c 6373 7464 696f #INCLUDE <CSTDIO 0010: 3e > 0000: 7072 696e 7466 2822 4865 6c6c 6f2c 2057 PRINTF("hELLO, w 0010: 6f72 6c64 215c 6e22 293b ORLD!\N"); 0000: 6d61 696e 203d 2064 6f20 6765 744c 696e MAIN = DO GETlIN 0010: 6520 3e3e 3d20 7072 696e 7420 2e20 7375 E >>= PRINT . SU 0020: 6d20 2e20 6d61 7020 7265 6164 202e 2077 M . MAP READ . W 0030: 6f72 6473 ORDS

 

Author

WU, Zejun

 

Source

2011 Asia Dalian Regional Contest

最后一个字符的转换完成后不需要补全，直接换行，注意边界条件的判断

#include<iostream>	
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn = 200010;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);

typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;

string tohex(char c)
{
	string ans = "00";
	int p = c, cnt = 1;
	while (p)
	{
		if (p % 16 < 10)
		{
			ans[cnt] = p % 16 + '0';
		}
		else
		{
			ans[cnt] = p % 16 - 10 + 'a';
		}
		cnt--;
		p /= 16;
	}
	return ans;
}

struct Line {
	string a;
	void operator++()
	{
		bool p = 1;
		int cur = 2;
		while (p)
		{
			p = 0;
			if (a[cur] != '9'&&a[cur] != 'f') a[cur]++;
			else
			{
				if (a[cur] == '9')
					a[cur] = 'a';
				else
				{
					a[cur] = '0';
					p = 1;
				}
			}
			cur--;
		}
	}
};

int main()
{
	string a;
	while (getline(cin, a))
	{
		Line line;
		line.a = "0000";
		int i = 0;
		for (i = 0; i < a.length(); i++)
		{
			if (i % 16 == 0)//判断是否换行
			{
				if (i != 0)
				{
					cout << endl;
				}
				cout << line.a << ": ";
				++line;
			}
			cout << tohex(a[i]);//前面输出
			if (i % 2 != 0)//判断前面的空格
			{
				cout << " ";
			}
			if (i&&(i+1) % 16 == 0 || i == a.length() - 1)
			{
				if ((i + 1) % 16 != 0) //补前面空格
				{
					for (int j = i % 16 + 1; j <= 15; j++)
					{
						cout << "  ";
						if (j % 2 != 0) cout << " ";
					}
				}
				for (int j = i - i % 16; j <= i; j++)//后面转换
				{
					if (a[j] >= 'A'&&a[j] <= 'Z')
					{
						cout << (char)(a[j] - 'A' + 'a');
					}
					else if (a[j] >= 'a'&&a[j] <= 'z')
					{
						cout << (char)(a[j] - 'a' + 'A');
					}
					else cout << a[j];
				}
			}
		}
		cout << endl;
	}
}

代码写的好丑，可以参照点击打开链接写的比较精简。