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hdu 1020 Encoding

2024-08-13 12:10:11   213人阅读

Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40214    Accepted Submission(s): 17846


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

 分析：有两种比较方式。第一种是s[i]与s[i+1]比较，当与s[i+1]不同时，则按要求输出s[i]，这里如果是定义string s,

最后s[s.length() - 1]和s[s.length()]比较会报错（用的是VS2010），所以我定义成char s[10001],最后s[strlen[s] - 1]和s[strlen[s]] = ‘\0‘比较没有报错。

 1 #include <iostream> 2 #include <cstring> 3 using namespace std; 4 int main(){ 5     int n; 6     char s[10001]; 7     cin >> n; 8     while(n--){ 9         cin >> s;10         int num = 1;11         int len = strlen(s);12         for(int i = 0; i < len; i++){13             if(s[i] == s[i+1])14                 num++;15             else {16                 if(num == 1){17                     cout << s[i];18                 } else {19                     cout << num << s[i];20                     num = 1;21                 }22             }23         }24         cout << endl;25     }26     return 0;27 }

第二种比较方法则是从下标1开始，s[i]和s[i-1]比较。当s[i] != s[i-1],则按要求输出s[i-1]。注意：不管s[s.length()-1]和s[s.length()-2]相不相等，都不会输出最后一个相同的字符字串。

 1 #include <iostream> 2 #include <string> 3 using namespace std; 4 int main(){ 5     int n; 6     string s; 7     cin >> n; 8     while(n--){ 9         cin >> s;10         int num = 1;11         int len = s.length();12         for(int i = 1; i < len; i++){13             if(s[i - 1] == s[i]){14                 num++;15             } else {16                 if(num != 1){17                     cout << num << s[i - 1];18                     num = 1;19                 }20                 else {21                     cout << s[i - 1];22                 }23             }24         }25         if(num != 1)26             cout <<  num << s[len - 1];27         else28             cout << s[len - 1];29         cout << endl;30     }31     return 0;32 }

 

hdu 1020 Encoding


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