本文主要是介绍HDU 4731 Minimum palindrome,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
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
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
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
Sample Input
2 2 2 2 3
Sample Output
Case #1: ab Case #2: aab
Source
Recommend
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;
}
这篇关于HDU 4731 Minimum palindrome的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!