HDU 4731 Minimum palindrome（规律 构造）

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

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 <= 10 ⁵)

 

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

转：

题意，用前m个字母，组成长度为n的字符串，使得子串中回文数最小
当m = 1时，必定全是a。
当m >= 3时，必定全是abc循环，因为要构成回文，回文中左右端点相同，且第二级相同。举例证明：
a的右边必定是b，a的左边必定是c。所以含a的必定不能成回文。而不含a的只有bc，所以abc循环回文数最小。
当m = 2时，思考用上述m>2的循环来接。当只有a，b两个字母。所以要两个字母组合，有aa,bb,ab,ba
任意取三个判断下，回文数最小，字典序最小。可以发现aababb循环的回文数最小，为4.当n<=8时需要特判，
因为n<=8时,最小回文数小于4.当n>=9时，开头添加aa不会增大回文数，且使字典序最小。删去aa后，接着就是
aababb循环了。
注意的是，aababb循环有剩余的时候，还需要特判。因为剩余的数不会有右边的数影响，可以用更小的字典序。 

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
    int t;
    int n, m;
    int cas = 0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&m,&n);
        printf("Case #%d: ",++cas);
        if(m == 1)
        {
            for(int i = 0; i < n; i++)
            {
                printf("a");
            }
        }
        else if(m == 2)
        {
            if(n == 1)
                printf("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");//首先输出aa
                n = n-2;
                int num = n/6;
                for(int i = 1; i <= num; i++)
                {
                    printf("aababb");
                }
                if(n%6 == 1)
                    printf("a");
                if(n%6 == 2)
                    printf("aa");
                if(n%6 == 3)
                    printf("aaa");
                if(n%6 == 4)
                    printf("aaaa");
                if(n%6 == 5)
                    printf("aabab");
            }
        }
        else
        {
            int num = n/3;
            for(int i = 1; i <= num; i++)
            {
                printf("abc");
            }
            if(n%3 == 1)
                printf("a");
            if(n%3 == 2)
                printf("ab");
        }
        printf("\n");
    }
    return 0;
}