Encoding （HDU 1020）

最新推荐文章于 2021-04-27 21:28:59 发布

叶发发

最新推荐文章于 2021-04-27 21:28:59 发布

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文章标签： HDU oj 字符串

本文链接：https://blog.youkuaiyun.com/weixin_42171092/article/details/82186227

博客围绕HDU oj上的字符串编码问题展开，给出问题描述、输入输出要求及示例。问题是对仅含‘A’ - ‘Z’的字符串编码，相同字符子串按规则转换。还提到易错点是要统计相邻字母间数字，最后给出参考代码。

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Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

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

这个题有个易错点：应该统计相邻两字母间的数字。而由于自己忽视了这一点，WA了好多次，下次要考虑周到些~~~
参考代码如下

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
    int n,i;
    char a[10001];
    scanf("%d",&n);
    while(n--)
    {
        scanf("%s",a);
        int num=1;
        for(i=0; a[i]!='\0'; i++)
        {
            if(a[i]==a[i+1]) num++;
            else
            {
                if(num>1)
                {
                    printf("%d%c",num,a[i]);
                    num=1;//一定要重置为1，切记！！！
                }
                else
                    printf("%c",a[i]);
            }
        }
        printf("\n");
    }



    return 0;
}