HDU 1020 Encoding【连续的计数器重置】
2024-08-30 00:43:52
Encoding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51785 Accepted Submission(s): 23041
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
ABC
ABBCCC
Sample Output
ABC
A2B3C
A2B3C
Author
ZHANG Zheng
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#include<map>
#include<sstream>
#include<queue>
#include<cmath>
#include<list>
#include<vector>
#include<string>
using namespace std;
#define long long ll
const double PI = acos(-1.0);
const double eps = 1e-;
const int inf = 0x3f3f3f3f;
const int N = ;
int n, m, tot;
int a[N]; map<string,int> mp;
int main()
{
int t;
cin >> t;
string s;
while(t--)
{
int c = ;
cin >> s;
for(int i=; i<s.size(); i++)
{
if(s[i+]==s[i]){
c++;
}
else{
if(c==){
printf("%c",s[i]);
c = ;
}
else{
printf("%d%c",c,s[i]);
c = ;
}
} }
cout<<endl;
}
}
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