Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 21463 Accepted Submission(s): 9352
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
2
ABC
ABBCCC
Sample Output
ABC
A2B3C
Author
ZHANG Zheng
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#include<stdio.h>
#include<string.h>
#define M 10000
int main()
{
int n;
char a[M+10];
scanf("%d",&n);
{
while(n--)
{
int i,j,l,k=0;
getchar();
scanf("%s",a);
l=strlen(a);
for(i=0;i<l;i++)
{
if(a[i]==a[i+1])
a[i]=0;
}
for(i=0;i<l;i++)
{
if(a[i]==0)
k++;
else
{
if(k==0)
printf("%c",a[i]);
else
printf("%d%c",k+1,a[i]);
k=0;
}
}
printf(" ");
}
}
return 0;
}
#include<string.h>
#define M 10000
int main()
{
int n;
char a[M+10];
scanf("%d",&n);
{
while(n--)
{
int i,j,l,k=0;
getchar();
scanf("%s",a);
l=strlen(a);
for(i=0;i<l;i++)
{
if(a[i]==a[i+1])
a[i]=0;
}
for(i=0;i<l;i++)
{
if(a[i]==0)
k++;
else
{
if(k==0)
printf("%c",a[i]);
else
printf("%d%c",k+1,a[i]);
k=0;
}
}
printf(" ");
}
}
return 0;
}
此题的k是连续字母的的个数。
代码如下：