这是 meelo 原创的 IEEEXtreme极限编程大赛题解
Xtreme 10.0 - N-Palindromes
题目来源 第10届IEEE极限编程大赛
https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/n-palindromes
Alice thinks that contest problem authors' obsession with palindromes is misplaced. She is much fonder of n-palindromes, which are words that are palindromes when the characters at exactly n positions are changed.
For example, Alice knows that her name (in lowercase) is a 2-palindrome, because she can create any of the following palindromes from her name by changing 2 characters: alila, acica, elile, ecice.
She also knows that her name is a 3-palindrome, because she can create palindromes by changing characters at 3 positions, e.g. ecace and zlilz. However, this is only a partial list, and she wants your help in determining the total number of such palindromes.
Note that the characters of an n-palindrome, including the n replacement characters, must all be lowercase English letters.
Input Format
The input starts with an integer t, on a line by itself, which gives the number of test cases.
Each test case is made up of an integer n followed by a lowercase string.
Constraints
1 ≤ t ≤ 20
1 ≤ n ≤ [length of string] ≤ 500
Output Format
For each test case, you should output, on a line by itself, the total number of palindromes that can be created by changing exactly n characters of the given string. Since this number may be very large, you should output the number modulo (109 + 7).
Sample Input
3
2 alice
1 racecar
3 alice
Sample Output
4
25
196
Explanation
The problem statement lists the four palindromes that can be made from the string alice, by changing 2 characters.
Since you can only change one character in racecar, you are constrained to changing the middle letter. This character can be changed to any of the 25 letters other than e.
For the last testcase, Alice has found that there are 196 palindromes that can be made from her name, by changing 3 characters.
题目解析
这是一道动态规划的题，动态规划的题一般都是要求一个超级大的数。
关键是状态怎么取。有3个关键的量决定结果的值。
可以修改的次数N，一定是一个决定状态的变量。因为可以修改的次数不同，结果一定不同。
一个长度为L的字符串，共有(L+1)//2对字符。如果所有的字母对都相同，则一定是回文串。既然只与一对字母有关，考虑问题的时候，总是考虑一对字母就好了。
共有(L+1)//2对字符，考虑一个子问题是从第1个字符算起的，前k对字符。这又是一个状态变量。
断定一个字符串是否是回文串，可以统计它不匹配的对数。如果不匹配的对数为0，则是回文串。
剩余的一个状态变量就是，不匹配的对数b了。
f(n,k,b)表示允许n次修改，只修改前k对字符，不匹配数为b时的回文串数。
状态转移方程为：
有三种情况：
第1种，最特殊的一种，长度为奇数的字符串，考虑正中央的字符。只有一个字符，一定是匹配的。可以选择修改，有25种修改的方法；或者不修改。
第2种，一对字符匹配。可以选择修改，同样有25种改法；或者不修改。
第3种，一对字符不匹配。一定要修改才能成为回文串。修改两个字符有24种改法；只该一个字符有两种改法，将一对字符的第2个改成第1个或者将第1个改成第2个。
考虑第3个样例，alice修改3次成为回文串。
f(3,3,2)=25f(2,2,2)+f(3,2,2)
=25(24f(0,1,1)+2f(1,1,1))+(24f(1,1,1)+2f(2,1,1))
=25(0+2(2f(0,0,0)))+(24(2f(0,0,0)+2(24f(0,0,0)))
=25*2*2+24*2+24*2
=196
程序
C++
#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> #include <string> using namespace std; #define ll long long #define MAX_LENGTH 510 #define MAX_N 1000000007 ll dp[MAX_LENGTH][MAX_LENGTH/2][MAX_LENGTH/2] = {0}; ll NPalindromes(const string &str, int N) { int length = str.length(); int num_group = (length + 1) / 2; // number of pair int num_mismatch = 0; for(int i=0; i<num_group; i++) { if(str[i] != str[length-i-1]) { num_mismatch++; } } dp[0][0][0] = 1; // change n character for(int n=0; n<=N; n++) { // only consider first k character for(int k=1; k<=num_group; k++) { // exist b mismatch for(int b=0; b<=num_mismatch; b++) { if(k-1 == length-k) { // odd string, considering middle character dp[n][k][b] = dp[n][k-1][b]; if(n>=1) dp[n][k][b] = (dp[n][k][b] + 25 * dp[n-1][k-1][b]) % MAX_N; } else if(str[k-1] == str[length-k]) { dp[n][k][b] = dp[n][k-1][b]; if(n>=2) dp[n][k][b] = (dp[n][k][b] + 25 * dp[n-2][k-1][b]) % MAX_N; } else if(str[k-1] != str[length-k]) { if(b>=1 && n>=1) dp[n][k][b] = 2 * dp[n-1][k-1][b-1]; if(n>=2) dp[n][k][b] = (dp[n][k][b] + 24 * dp[n-2][k-1][b-1]) % MAX_N; } } } } return dp[N][num_group][num_mismatch]; } int main() { int T; cin >> T; string str; int n; for(int t=0; t<T; t++) { cin >> n >> str; cout << NPalindromes(str, n) << endl; } return 0; }