HDU-1020-Encoding

最新推荐文章于 2024-07-21 10:00:00 发布

Prim233

最新推荐文章于 2024-07-21 10:00:00 发布

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分类专栏：字符串语言入门文章标签： string STL

本文链接：https://blog.youkuaiyun.com/qq_32680617/article/details/51124799

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本文介绍了一种简单的字符串压缩算法，该算法将连续重复的字符替换为字符及其出现次数。通过示例展示了输入输出过程，并提供了完整的C++实现代码。

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Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38191 Accepted Submission(s): 16932

Problem Description
Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:

Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.
If the length of the sub-string is 1, ‘1’ should be ignored.

Input
The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only ‘A’ - ‘Z’ and the length is less than 10000.

Output
For each test case, output the encoded string in a line.

Sample Input
2
ABC
ABBCCC

Sample Output
ABC
A2B3C

题意不懂？看样例！
string水题，就为了练习一下string
string用法http://blog.youkuaiyun.com/qq_32680617/article/details/51122395

代码

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        string s1;
        cin>>s1;
        int num=0;
        for(string::size_type i=0; i<s1.size(); i++)
        {
            if(s1[i]==s1[i+1])
            {
                num=1;
                while(s1[i]==s1[i+1]&&i<s1.size()-1)
                {
                    num++;
                    i++;
                }
                cout<<num<<s1[i-1];
            }
            else
                cout<<s1[i];
        }
        cout<<endl;
    }
    return 0;
}

再水几道，熟悉一下string