[HDU 1020] Encoding

Encoding

 

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/32768 K (Java/Others)

 

 

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

 

 

AC Code

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
 
#define MAXLEN 10010
 
 
char * Encording(const char * pszBuf)
{
    int iNum = 0;
    int iNumLen = 0;
    int iLen = 0;
    int i = 0;
    int j = 0;
    int k = 0;
    char * pszResult = NULL;
 
    assert(pszBuf != NULL);
 
    iLen = strlen(pszBuf);
 
    pszResult = (char *)malloc(sizeof(char)* (iLen + 1));
 
    if (NULL == pszResult)
    {
        return NULL;
    }
 
    memset(pszResult, 0, sizeof(char)* (iLen + 1));
 
    for (i = 0; i < iLen; i++)
    {
        iNum = 0;
        for (j = i + 1; j < iLen; j++)
        {
            if (*(pszBuf + i) == *(pszBuf + j))
            {
                iNum = j - i + 1;
                continue;
            }
            else
            {
                break;
            }
        }
 
        if (iNum > 1)
        {
            iNumLen = sprintf(pszResult + k, "%d", iNum);
            *(pszResult + k + iNumLen) = *(pszBuf + i);
            k += iNumLen + 1;
            i += iNum - 1;
        }
        else
        {
            *(pszResult + k) = *(pszBuf + i);
            k++;
        }
    }
 
    return pszResult;
}
 
 
#ifdef ONLINE_JUDGE
int main()
{
    int iTotalLine = 0;
    char *pszTmp = NULL;
    char *pszResult = NULL;
 
    pszTmp = (char *)malloc(sizeof(char)* (MAXLEN + 1));
 
    if (NULL == pszTmp)
    {
        return -1;
    }
 
    while (scanf("%d", &iTotalLine) != EOF)
    {
        assert((1 <= iTotalLine) && (iTotalLine <= 100));
        while (iTotalLine--)
        {
            memset(pszTmp, 0, sizeof(char)* (MAXLEN + 1));
 
            scanf("%s", pszTmp);
            pszResult = Encording(pszTmp);
            if (NULL == pszResult)
            {
                free(pszTmp);
                pszTmp = NULL;
                return -1;
            }
            printf("%s
", pszResult);
 
            free(pszResult);
            pszResult = NULL;
        }
    }
 
    free(pszTmp);
    pszTmp = NULL;
 
    assert(pszResult == NULL);
 
    return 0;
}
#endif

TestCase

class OJTest : public testing::Test
{
protected:
    virtual void SetUp()
    {
        m_pszResult = NULL;
    }
    virtual void TearDown()
    {
        free(m_pszResult);
        m_pszResult = NULL;
    }
protected:
    char * m_pszResult;
};
 
TEST_F(OJTest, TestCase01)
{
    m_pszResult = Encording("ABC");
    ASSERT_STREQ("ABC", m_pszResult);
}
 
TEST_F(OJTest, TestCase02)
{
    m_pszResult = Encording("ABBCCC");
    ASSERT_STREQ("A2B3C", m_pszResult);
}
 
TEST_F(OJTest, TestCase03)
{
    m_pszResult = Encording("ABBCBBAACC");
    ASSERT_STREQ("A2BC2B2A2C", m_pszResult);
}
 
TEST_F(OJTest, TestCase04)
{
    m_pszResult = Encording("AAAAAAAAAAABC");
    ASSERT_STREQ("11ABC", m_pszResult);
}