hdu-1020 Encoding

最新推荐文章于 2024-07-21 10:00:00 发布

转载最新推荐文章于 2024-07-21 10:00:00 发布 · 85 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/love-sherry/p/6941430.html

文章标签：

#java #php

本文介绍了一种字符串处理算法，该算法能将输入字符串中连续重复的字符进行缩略编码，如“ABBCCC”被编码为“A2B3C”。文章提供了一个具体的编程实现案例，包括完整的C语言代码示例，可用于理解如何实现字符串的缩略。

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1020

题目类型：

字符串的简单处理

题意概括：

对字符串进行缩略，将一个字符串中连续的n个相同的字符缩略成字符n的形式。

题目：

Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45946 Accepted Submission(s): 20310

Problem Description

Given a string containing only 'A' - 'Z', we could encode it using the following method:

1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.

2. 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

ABC

ABBCCC

Sample Output

ABC

A2B3C

# include <stdio.h>
# include <string.h>

int main ()
{
    int i,j,t,l;
    char b[10010];
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        gets(b);
        l=strlen(b);
        b[l]='a';b[l+1]='\0';
        l++;
        int num=1;
        for(i=1;i<l;i++)
        {
            if(b[i]!=b[i-1])
            {
                if(num!=1)
                    printf("%d",num);
                printf("%c",b[i-1]);
                num=1;
            }
            else
                num++;
        }
        printf("\n");
    }
}