Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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
AC Code --Java
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
while (n-- != 0) {
String a = sc.next();
char[] arr = a.toCharArray();
char[] zc = new char[10000];
int[] zs = new int[5000];
int j = 0;
zc[0] = arr[0];
for (int i = 1; i < arr.length; i++) {
if (arr[i] == arr[i - 1]) {
zs[j]++;
} else {
zc[++j] = arr[i];
}
}
for (int i = 0; i < j + 1; i++) {
if (zs[i] == 0)
System.out.print(zc[i]);
else
System.out.print(zs[i] + 1 + "" + zc[i]);
}
System.out.println();
}
}
}
C 代码版
#include <stdio.h> void out(char arr[100]);
int main(void)
{ int n=0,i=0;
char arr1[100][100];
while(scanf("%d",&n)!=EOF)
{
getchar();
for(i=0;i<n;i++)
{
gets(arr1[i]);
} for(i=0;i<n;i++)
{
out(arr1[i]);
}
} return 0;
}
void out(char arr[100])
{
char s[100];
int sum[100];
int i=0,j=0;
s[0]=arr[0];
sum[0]=1;
for(i=1;i<strlen(arr)+1;i++)
{
if(arr[i]==arr[i-1])
sum[j]++;
else
{
s[++j]=arr[i];
sum[j]=1;
}
}
s[j]='\0';
for(i=0;i<strlen(s);i++)
if(sum[i]!=1)
printf("%d%c",sum[i],s[i]);
else
printf("%c",s[i]);
printf("\n");
}