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Encoding

时间:2015-01-12 10:58:17      阅读:135      评论:0      收藏:0      [点我收藏+]

标签:printf   数据   测试   

Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28533 Accepted Submission(s): 12642


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
源代码一:考虑字符串数组中的字符结尾
#include <stdio.h> #include <string.h> #include <stdlib.h> char stra[10000]; int main() {   int i,j,n,lena,count;   scanf("%d",&n);   getchar();   while(n--)   {      scanf("%s",stra);     count=1;     lena=strlen(stra);     for(i=0,j=0;i<lena-1;i++)       if(stra[i]==stra[i+1])         count++;        else       {         if(count==1)           printf("%c",stra[i]);           else          {           printf("%d%c",count,stra[i]);             count=1;          }          }     if((i==lena-1) && count>1)        printf("%d",count);          printf("%c\n",stra[i]);         }   system("pause");   return 0;     }
源代码二:直接使用字符数组
#include <stdio.h> #include <string.h> #include <stdlib.h> char a[10001]; int main() { int n,len,i,count; scanf("%d",&n); getchar(); while(n--) { count=1; gets(a); len=strlen(a); for(i=0;i<len;i++) { if(a[i]==a[i+1]) count++; else { if(count==1) printf("%c",a[i]); else { printf("%d%c",count,a[i]); count=1; } }
} printf("\n"); } system("pause"); return 0; }

Encoding

标签:printf   数据   测试   

原文地址:http://blog.csdn.net/zchlww/article/details/42639089

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