• 杭电1020Encoding


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

    题目:

    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
     
    直接扫描一遍就好,不用用数组储存数据;
     1 #include <iostream>
     2 #include <algorithm>
     3 #include <cstdio>
     4 #include <cmath>
     5 #include <cstring>
     6 #include <queue>
     7 #include <stack>
     8 #include <map>
     9 #include <vector>
    10 
    11 #define PI acos((double)-1)
    12 #define E exp(double(1))
    13 using namespace std;
    14 
    15 int main (void)
    16 {
    17     int t,n;
    18     char a,b;
    19     cin>>t;
    20     scanf("%*c");
    21     while(t--)
    22     {
    23         n=1;
    24         scanf("%c",&a);
    25         while(scanf("%c",&b)==1 && b!='
    ')
    26         {
    27             if(a == b)
    28             {
    29                     n++;
    30             }
    31             else
    32             {
    33                 if(n>1)
    34                     printf("%d%c",n,a);
    35                 else
    36                     printf("%c",a);
    37                 n=1;
    38                 a=b;
    39             }
    40         }
    41         if(n>1)
    42                 printf("%d%c
    ",n,a);
    43             else
    44                 printf("%c
    ",a);
    45     }
    46     return 0;
    47 }
    View Code

    作者:weeping
    出处:www.cnblogs.com/weeping/
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。

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  • 原文地址:https://www.cnblogs.com/weeping/p/5351848.html
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