Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29834 Accepted 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
2 ABC ABBCCC
Sample Output
ABC A2B3C
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]; else cout<<count<<str[i-1]; count=1; } } if(count>1) cout<<count<<str[i-1]; cout<<endl; } return 0; }