pid=1020">Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25691    Accepted Submission(s): 11289 Problem Description Given a string containing only 'A' - 'Z', we could encode it usi…

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u 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 &quo…

Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11132    Accepted Submission(s): 4673 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followin…

//自信满满地交上去~~but...超时了 #include <iostream> #include <string.h> #include <stdio.h> using namespace std; int main() { ],c; int i,k,j; cin>>i; while(i) { cin>>ch; ;k<strlen(ch);) { ]) {cout<<ch[k];k++;} else { c=ch[k]; f…

题意是要统计在一段字符串中连续相同的字符,不用再排序,相等但不连续的字符要分开输出,不用合在一起,之前用了桶排序的方法一直 wa,想复杂了. 代码如下: #include <bits/stdc++.h> using namespace std; int main() { std::ios::sync_with_stdio(false); int t,num,len; char c; bool f; string s; cin >> t; while(t--) { cin >&…

Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39047    Accepted Submission(s): 17279 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followi…

Encoding Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 55351    Accepted Submission(s): 24697 Problem Description Given a string containing only 'A' - 'Z', we could encode it using the followi…

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 followi…

并不是很精简,随便改改A过了就没有再简化了. 1020. 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 c…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…