hdu Encoding
1782 ワード
Encoding
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 16 Accepted Submission(s) : 13
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
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 16 Accepted Submission(s) : 13
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
#include <stdio.h>
#include <string.h>
int main()
{
char str[10005];
int N,i,j=0,k;
scanf("%d",&N);
getchar(); //
while (N--)
{
memset(str,0,sizeof(str));
// getchar(); , , !
gets(str);
int len=0;
len=strlen(str);
k=1;
for (i=0;i<len;i++)
{
if (str[i+1]==str[i])// , 0,0
{
k++;
}
else
{
if (k!=1) printf("%d",k);
printf("%c",str[i]);
k=1; //k 1
}
}
printf("
");
}
return 0;
}