题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4731

Minimum palindrome

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1439    Accepted Submission(s): 600

Problem Description

Setting password is very important, especially when you have so many "interesting" things in "F:\TDDOWNLOAD". We define the safety of a password by a value. First, we find all the substrings of the password. Then we calculate the maximum length of those substrings which, at the meantime, is a palindrome. A palindrome is a string that will be the same when writing backwards. For example, aba, abba,abcba are all palindromes, but abcab, abab are not. A substring of S is a continous string cut from S. bcd, cd are the substrings of abcde, but acd,ce are not. Note that abcde is also the substring of abcde. The smaller the value is, the safer the password will be. You want to set your password using the first M letters from the alphabet, and its length should be N. Output a password with the smallest value. If there are multiple solutions, output the lexicographically smallest one. All the letters are lowercase.  

Input

The first line has a number T (T <= 15) , indicating the number of test cases. For each test case, there is a single line with two integers M and N, as described above.(1 <= M <= 26, 1 <= N <= 10⁵)  

Output

For test case X, output "Case #X: " first, then output the best password.  

Sample Input

2 2 2 2 3  

Sample Output

Case #1: ab Case #2: aab  

Source

2013 ACM/ICPC Asia Regional Chengdu Online  

Recommend

liuyiding

思路：就是找规律。尼玛，比赛时以为M个字母全要用上，吐血啊！当M=1时，肯定全是a；当M>2时，是abc循环使用；困难的是M=2时，其实写个暴力程序打表出来，然后找规律即可。如下所示。

打表代码如下：

#include <bits/stdc++.h> using namespace std; int num[40]; int main(){ ios::sync_with_stdio(false); cin.tie(0); for (int i=1; i<=25; ++i){ int maxid = i; int ans = 0; for (int j=0; j<(1<<i); ++j){ int t = j; for (int k=0; k<i; ++k){ num[k] = t&1; t >>= 1; } int minid = 0; for (int k=0; k<i; ++k){ for (int s=0; k-s>=0&&k+s<i; ++s){ if (num[k-s] != num[k+s]) break; if (s*2+1 > minid) minid = s*2+1; } for (int s=0; k-s>=0&&k+s+1<i; ++s){ if (num[k-s] != num[k+s+1]) break; if (s*2+2 > minid) minid = s*2+2; } } if (minid < maxid){ maxid = minid; ans = j; } } for (int j=i-1; j>=0; --j){ num[j] = ans&1; ans >>= 1; } cout << i << ": "; for (int j=0; j<i; ++j) cout << char(num[j]+'a'); cout << endl; } return 0; } 附上AC代码：

#include <bits/stdc++.h> using namespace std; int n, m; int main(){ int T, cas=0; scanf("%d", &T); while (T--){ scanf("%d%d", &m, &n); printf("Case #%d: ", ++cas); if (m == 1) while (n--) putchar('a'); else if (m == 2){ if (n == 1) putchar('a'); else if (n == 2) printf("ab"); else if (n == 3) printf("aab"); else if (n == 4) printf("aabb"); else if (n == 5) printf("aaaba"); else if (n == 6) printf("aaabab"); else if (n == 7) printf("aaababb"); else if (n == 8) printf("aaababbb"); else{ printf("aa"); int t = (n-2)/6; while (t--) printf("aababb"); t = (n-2)%6; if (t == 1) putchar('a'); else if (t == 2) printf("aa"); else if (t == 3) printf("aaa"); else if (t == 4) printf("aaaa"); else if (t == 5) printf("aabab"); } } else{ int t = n/3; while (t--) printf("abc"); t = n%3; if (t == 1) putchar('a'); else if (t == 2) printf("ab"); } puts(""); } return 0; }