XDOJ1207 - love message

最新推荐文章于 2022-04-02 16:36:19 发布

原创最新推荐文章于 2022-04-02 16:36:19 发布 · 716 阅读

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本文介绍了一段关于使用特定算法加密和解密恋爱间秘密信息的故事。通过将字符出现次数转化为特定表达，江山和赵怡然在不同班级中巧妙地传递爱情信息，然而他们的老师却能截获所有信息。为了保护爱情的秘密，他们需要一种方法来破解信息。本文提供了解决方案，并通过简单的字符串处理展示了如何实现这一目标。

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Description

Jiangshan and Zhao Yiran is a sweetheart lover in different class. So, Jiangshan developed a application to send message to Zhao Yiran through wireless. However, so unfortunately, their teachers are so great that all they send will be received by their teachers too. In order to keep their message secrecy, they have to encrypt the message.

When a char X appear n times, Jiangshan will show it as Xn, for example bb->b2. In order to get the love message, Zhao Yiran have to deciphering the message with mind acts upon mind. Could you help they to deciphering.

Input

A string S, indicates the message Zhao Yiran get, all the char is lower case. There may be several test cases in the input file and end when S equal to XXX.

Output

A string T, indicates the message after deciphering.

Sample Input

a3
go12gle
XXX

Sample Output

aaa
goooooooooooogle

解题思路：

非常简单的字符串处理

#include<iostream>
#include<string>
using namespace std;

int main()
{
    string str;
    string over = "XXX";
    string ans;
    while(cin>>str)
    {
        if(str==over)
            break;
        int len = str.length();
        ans = "";
        for(int i=0;i<len;)
        {
            char ch;
            if(str[i]>='0'&&str[i]<='9')
            {
                ch = str[i-1];
                int t = str[i]-'0';
                ++i;
                while(str[i]>='0'&&str[i]<='9')
                {
                    t *= 10;
                    t += str[i]-'0';
                    ++i;
                }
                for(int j=1;j<t;++j)
                    ans += ch;
            }
            else
            {
                ans += str[i];
                ++i;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

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