ACM趣题天天练 6 Encoding
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 100.
Output
For each test case, output the encoded string in a line.
Sample Input
2
ABC
ABBCCC
Sample Output
ABC
A2B3C
来自：http://acm.zju.edu.cn/2478
#include <iostream>
#include <string>
using namespace std;
int main()
{
int n,i,len,k;
string str;
char tmp;
while(cin >> n)
{
while(n--)
{
cin >> str;
len = str.length();
for(i = 0; i < len; ++i)
{
tmp = str;
k = 1;
while(str[++i] == tmp)
{
k++;
}
if(k != 1)
{
cout << k;
}
cout << tmp;
i--;
}
cout << endl;
}
}
return 0;
}
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = Integer.parseInt(sc.nextLine());
while (n-- > 0) {
int count;
char temp;
String str = sc.nextLine();
for (int i = 0; i < str.length(); i++) {
count = 0;
temp = str.charAt(i);
while (i < str.length() && str.charAt(i) == temp) {
count++;
i++;
}
if (count > 1) {
System.out.print(count);
}
System.out.print(temp);
i--;
}
System.out.println();
}
sc.close();
}
}
n = input()
for a in range(n):
s = raw_input() + '\0'
res = ""
ch = 0
n = 0
for c in s:
if c == ch:
n += 1
elif c != ch:
if n == 1:
res += ch
elif n > 1:
res += str(n) + ch
ch = c
n = 1
print res
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