hdu1020
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Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11421 Accepted Submission(s): 4802
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.
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
2ABCABBCCC
Sample Output
ABCA2B3C
#include<stdio.h>
#include<string.h>
char str[10009];
int main()
{
int nCases,i,len,k;
char c;
scanf("%d",&nCases);
while(nCases--)
{
//getchar();
scanf("%s",str);
i=0;len=strlen(str);
while(i<len)
{
c=str[i];
k=i;//k记下当前位置,i向后移动
while(str[i]==c&&i<len)
i++;
if(i-k>1)
printf("%d%c",i-k,c);
else
printf("%c",c);
}
printf("\n");
}
return 0;
}
#include<string.h>
char str[10009];
int main()
{
int nCases,i,len,k;
char c;
scanf("%d",&nCases);
while(nCases--)
{
//getchar();
scanf("%s",str);
i=0;len=strlen(str);
while(i<len)
{
c=str[i];
k=i;//k记下当前位置,i向后移动
while(str[i]==c&&i<len)
i++;
if(i-k>1)
printf("%d%c",i-k,c);
else
printf("%c",c);
}
printf("\n");
}
return 0;
}
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