ACM题目 - 杭电1020 编码

最新推荐文章于 2019-08-11 17:40:01 发布

原创最新推荐文章于 2019-08-11 17:40:01 发布 · 597 阅读

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本文介绍了一种简单的字符串压缩编码算法，该算法通过识别并压缩连续重复的字符来减少字符串的长度。具体实现方法为：遇到连续重复的字符时，用重复次数加上字符本身来代替这些重复字符。例如，字符串“ABBCCC”会被压缩成“A2B3C”。文章提供了算法的详细描述及其实现代码。

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

#include<iostream>
#include<string>
using namespace std;
main(){
int n;
cin>>n;
while(n--){
string a;
cin>>a;
int sum=1;
int t=0,flag=0,x=1;
for(int i=0;i<a.length()-1;i++){
if(a[i]==a[i+1]){
a[i]='1';
}
}
for(int i=0;i<a.length();i++){
if(a[i]!='1' ){
if(sum!=1){
cout<<sum;
}
cout<<a[i];
sum=1;
}
else {
sum++;
}
}
cout<<endl;
}
}