HDU——1020 Encoding
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Encoding
*Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47730 Accepted Submission(s): 21182*
Problem Description
Given a string containing only ‘A’ - ‘Z’, we could encode it using the following method:
Each sub-string containing k same characters should be encoded to “kX” where “X” is the only character in this sub-string.
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 Input2ABCABBCCC
Sample OutputABCA2B3C
题目比较简单,思路就是以当前第一个与前面字符不相同的字符为基准,向后比较,找到第一个与它不同的字符。
C++代码:
#include <cstdio>#include <cstring>char str[10001];//像这种比较大的数组定义在主函数外面比较好,有时候定义在里面会引起不必要的错误int main(){ int N,len=0,cur,com,temp; scanf("%d",&N); while(N-- != 0){ scanf("%s",str); len = strlen(str); cur = 0; com = 0; while(cur < len && com < len){ while(str[com] == str[cur] && com < len) com ++; temp = com - cur; if(temp > 1) printf("%d",temp); printf("%c",str[cur]); cur = com; } printf("\n"); }}
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