hdu 1020 Encoding

Encoding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40214    Accepted Submission(s): 17846

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

 分析：有两种比较方式。第一种是s[i]与s[i+1]比较，当与s[i+1]不同时，则按要求输出s[i]，这里如果是定义string s,

最后s[s.length() - 1]和s[s.length()]比较会报错（用的是VS2010），所以我定义成char s[10001],最后s[strlen[s] - 1]和s[strlen[s]] = '\0'比较没有报错。

#include <iostream>
#include <cstring>
using namespace std;
int main(){
    int n;
    char s[10001];
    cin >> n;
    while(n--){
        cin >> s;
        int num = 1;
        int len = strlen(s);
        for(int i = 0; i < len; i++){
            if(s[i] == s[i+1])
                num++;
            else {
                if(num == 1){
                    cout << s[i];
                } else {
                    cout << num << s[i];
                    num = 1;
                }
            }
        }
        cout << endl;
    }
    return 0;
}

第二种比较方法则是从下标1开始，s[i]和s[i-1]比较。当s[i] != s[i-1],则按要求输出s[i-1]。注意：不管s[s.length()-1]和s[s.length()-2]相不相等，都不会输出最后一个相同的字符字串。

#include <iostream>
#include <string>
using namespace std;
int main(){
    int n;
    string s;
    cin >> n;
    while(n--){
        cin >> s;
        int num = 1;
        int len = s.length();
        for(int i = 1; i < len; i++){
            if(s[i - 1] == s[i]){
                num++;
            } else {
                if(num != 1){
                    cout << num << s[i - 1];
                    num = 1;
                }
                else {
                    cout << s[i - 1];
                }
            }
        }
        if(num != 1)
            cout <<  num << s[len - 1];
        else
            cout << s[len - 1];
        cout << endl;
    }
    return 0;
}