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  • HDU 1020 Encoding POJ 3438 Look and Say

    Encoding

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 11132    Accepted Submission(s): 4673

    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
      
    Author
    ZHANG Zheng
      
    Recommend
    JGShining
     
     1 #include<iostream>
     2 using namespace std;
     3 
     4 int main(){  
     5     int t;  
     6     char str[10001];  
     7     scanf("%d",&t);  
     8     while(t--){  
     9         scanf("%s",str);  
    10         int count=1;  
    11         for(int i=0;i<strlen(str);i++){  
    12             if(str[i]==str[i+1])count++;  
    13             else{  
    14                 if(count==1) printf("%c",str[i]);  
    15                 else printf("%d%c",count,str[i]);  
    16                 count=1;  
    17             }  
    18         }  
    19         printf("
    ");  
    20     }  
    21 } 
    View Code
     该题同poj中3438题
    http://poj.org/problem?id=3438
    Look and Say
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 9179   Accepted: 5552

    Description

    The look and say sequence is defined as follows. Start with any string of digits as the first element in the sequence. Each subsequent element is defined from the previous one by "verbally" describing the previous element. For example, the string 122344111 can be described as "one 1, two 2's, one 3, two 4's, three 1's". Therefore, the element that comes after 122344111 in the sequence is 1122132431. Similarly, the string 101 comes after 1111111111. Notice that it is generally not possible to uniquely identify the previous element of a particular element. For example, a string of 112213243 1's also yields 1122132431 as the next element.

    Input

    The input consists of a number of cases. The first line gives the number of cases to follow. Each case consists of a line of up to 1000 digits.

    Output

    For each test case, print the string that follows the given string.

    Sample Input

    3
    122344111
    1111111111
    12345

    Sample Output

    1122132431
    101
    1112131415

    Source

     1 #include<iostream>
     2 using namespace std;
     3 
     4 int main(){  
     5     int t;  
     6     char str[1001];  
     7     scanf("%d",&t);  
     8     while(t--){  
     9         scanf("%s",str);  
    10         int count=1;  
    11         for(int i=0;i<strlen(str);i++){  
    12             if(str[i]==str[i+1])count++;  
    13             else{  
    14                 printf("%d%c",count,str[i]);  
    15                 count=1;  
    16             }  
    17         }  
    18         printf("
    ");  
    19     }  
    20 }
    View Code
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  • 原文地址:https://www.cnblogs.com/samjustin/p/4567394.html
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