EncodingTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29834Accepted Submission(s): 13212
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
2ABCABBCCC
Sample Output
ABCA2B3C
Author
ZHANG Zheng
一个笔题 硬是他妈的让我wa次！错误原因是逻辑错误，如下错误代码：
虽然用一个数组记录下对应的字母出现次数；
但是输出的时候可是从他妈零依次检索，所以输出顺序完全就不同了但计数没问题；
#include<iostream>#include<string>using namespace std;int main(){string str;int s[10003];int N;cin>>N;while(N–){int ac=0;char ch;memset(s,0,sizeof(s));cin>>str;for(int i=0;i<str.size();i++){ac=str[i]-65;s[ac]++;}for(int j=0;j<26;j++)if(s[j]==1){ch=j+65;cout<<ch;}else if(s[j]!=1&&s[j]){ch=j+65;cout<<s[j]<<ch;}cout<<endl;}return 0;}
改正后，重新编写，once AC！：
#include<iostream>using namespace std;#include<string> int main(){string str;int n,i;cin>>n;while(n–){int count =1;cin>>str;for(i=1;i<str.size();i++){if(str[i]==str[i-1])count++;else{if(count==1)cout<<str[i-1];elsecout<<count<<str[i-1];count=1;}}if(count>1)cout<<count<<str[i-1];cout<<endl;}return 0;}
，欲望以提升热忱，毅力以磨平高山。
