DSP

POJ 1782 Run Length Encoding

2019-07-13 12:48发布

Run Length Encoding Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 4330   Accepted: 1405 Description Your task is to write a program that performs a simple form of run-length encoding, as described by the rules below. 

Any sequence of between 2 to 9 identical characters is encoded by two characters. The first character is the length of the sequence, represented by one of the characters 2 through 9. The second character is the value of the repeated character. A sequence of more than 9 identical characters is dealt with by first encoding 9 characters, then the remaining ones. 

Any sequence of characters that does not contain consecutive repetitions of any characters is represented by a 1 character followed by the sequence of characters, terminated with another 1. If a 1 appears as part of the sequence, it is escaped with a 1, thus two 1 characters are output. 
Input The input consists of letters (both upper- and lower-case), digits, spaces, and punctuation. Every line is terminated with a newline character and no other characters appear in the input. Output Each line in the input is encoded separately as described above. The newline at the end of each line is not encoded, but is passed directly to the output. Sample Input AAAAAABCCCC 12344 Sample Output 6A1B14C 11123124 Source Ulm Local 2004 PS 空格和空行也是元素 AC代码: #include #include #include using namespace std; int main(){ char s[1000]; while(gets(s)){ for(int i=0;s[i];++i){ if(s[i+1]!=s[i]&&s[i]){ printf("1"); while(s[i+1]!=s[i]&&s[i]){ putchar(s[i]); if(s[i]=='1') putchar(s[i]); i++; } i--; putchar('1'); } else if(s[i+1]==s[i]&&s[i]){ int cns=2; i++; while(s[i+1]==s[i]&&s[i]){ cns++; if(cns>9){ cns=9; break; } i++; } putchar(cns+'0'); putchar(s[i]); } } putchar(' '); } return 0; }