[HDOJ 1020]Encoding 字符串编码

                                       [HDOJ
1020]Encoding 字符串编码

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
思路： 一开始以为是输出所有的相同的，所以WA了两次，原来是子串的相同，要临近的数才能组合一起。。。。。我的代码：

#include <iostream>
#include <cstring>
using namespace std;
int main()
{
int i,k,n,count;
char a[10010];
cin >> n;
while (n--)
{
cin >> a;
k=strlen(a);
if (k==1)
{
cout << a << endl;
continue;
}
for (i=0,count=1;i<k;++i)
{
if (a[i]!=a[i+1])
{
if (count!=1)
{
cout << count ;
count=1;
}
cout << a[i];
}
else
{
count++;
}
}
cout << endl;
}
return 0;
}

网上的代码：

#include <stdio.h>
#include <string.h>
int main(){
int n,i,num;
char str[10001];//题目要求是10000个字符，我们要开大一点
scanf("%d",&n);
while(n--){
num=1;
scanf("%s",str);
for(i=0;i<strlen(str);i++){
if(str[i]==str[i+1]){ //如果一个字符跟它后面的字符相同则num++;
num++;
}
else{
if(num<=1) {
printf("%c",str[i]);
num=1;}//记住num要重置为1，不然残留后会导致后面出现错误
else{
printf("%d%c",num,str[i]);
num=1;}
}
}printf("\n");
}return 0;
}