UVA 458 The Decoder 简单解密

最新推荐文章于 2019-04-18 14:38:45 发布

小盼你最萌哒

最新推荐文章于 2019-04-18 14:38:45 发布

阅读量708

点赞数

CC 4.0 BY-SA版权

分类专栏：简单题

本文链接：https://blog.youkuaiyun.com/qq_27508477/article/details/47132107

简单题专栏收录该内容

14 篇文章

订阅专栏

本文介绍了一个简单的字符解码程序设计，该程序能够正确地将一组编码字符转换为有效的消息。通过对ASCII码进行特定的数学操作实现字符替换。文章提供了完整的程序代码及运行示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

原题： https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=399

题目：

The Decoder

Write a complete program that will correctly decode a set of characters into a valid message. Your program should read a given file of a simple coded set of characters and print the exact message that the characters contain. The code key for this simple coding is a one for one character substitution based upon a single arithmetic manipulation of the printable portion of the ASCII character set.

Input and Output

For example: with the input file that contains:

1JKJ’pz’{ol’{yhklthyr’vm’{ol’Jvu{yvs’Kh{h’Jvywvyh{pvu5
1PIT’pz’h’{yhklthyr’vm’{ol’Pu{lyuh{pvuhs’I|zpulzz’Thjopul’Jvywvyh{pvu5
1KLJ’pz’{ol’{yhklthyr’vm’{ol’Kpnp{hs’Lx|pwtlu{‘Jvywvyh{pvu5
your program should print the message:

*CDC is the trademark of the Control Data Corporation.
*IBM is a trademark of the International Business Machine Corporation.
*DEC is the trademark of the Digital Equipment Corporation.
Your program should accept all sets of characters that use the same encoding scheme and should print the actual message of each set of characters.

Sample Input

*CDC is the trademark of the Control Data Corporation.
*IBM is a trademark of the International Business Machine Corporation.
*DEC is the trademark of the Digital Equipment Corporation.

思路：

对于每一个字符，它变成Ascii减去一个常数的字符。讲一个字符串变成另一个字符串。

通过样例我们可以发现，任取两个字符，他们的ascii码的差值为7，所以只需要对读入的每个字符做一次减法即可。

代码：

#include <iostream>
#include"string.h"
#include"cstdio"
#include"stdlib.h"
#include"algorithm"
using namespace std;


int main()
{
    char s[1000];
    while(scanf("%s",s)!=EOF)
    {
        int l=strlen(s);
        for(int i=0;i<l;i++)
        {
            s[i]=s[i]-7;
        }
        printf("%s\n",s);
    }
    return 0;
}