HDU 1020 Encoding【连续的计数器重置】
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.
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
#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-6; const int inf = 0x3f3f3f3f; const int N = 500005; int n, m, tot; int a[N]; map<string,int> mp; int main() { int t; cin >> t; string s; while(t--) { int c = 1; cin >> s; for(int i=0; i<s.size(); i++) { if(s[i+1]==s[i]){ c++; } else{ if(c==1){ printf("%c",s[i]); c = 1; } else{ printf("%d%c",c,s[i]); c = 1; } } } cout<<endl; } }