HDU 1020 Encoding Encoding

95 阅读 0 评论 63 点赞

我是靠谱客的博主无奈唇彩，最近开发中收集的这篇文章主要介绍HDU 1020 Encoding Encoding，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28510 Accepted Submission(s): 12635

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

  
  
   
   2
ABC
ABBCCC

Sample Output

  
  
   
   ABC
A2B3C
  
  

水题 不说话
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

char s[10005];
struct point
{
  char c;
  int count;
}a[10005];

int main(void)
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
            scanf("%s",s);
            int l=strlen(s);
            int p=1;

            a[p].count=1;
            a[p].c=s[0];
            for(int i=1;i<l;i++)
            {
                if(s[i]!=s[i-1])
                {
                    a[++p].count=1;
                    a[p].c=s[i];
                }
                else a[p].count++;
            }
            for(int i=1;i<=p;i++)
            {
                if(a[i].count==1)
                    printf("%c",a[i].c);
                else
                    printf("%d%c",a[i].count,a[i].c);
            }
            printf("n");

        }
    }
    return 0;
}