B的三种做法
定义一个字符串的“伪回文值”是:修改最少字符数量使得其变成回文串的修改次数。例如,"abca"的伪回文值是1。任何回文串的伪回文值是0。
给定一个字符串,求出所有连续子串的伪回文值之和。(长度不超过100)
枚举|区间dp
法一:枚举所有字串,用双指针从两边判断需要修改多少个字符才能变为回文串。
法二:
:表示
字串的伪回文值
若
否则,
枚举所有字串,求和得答案。
法三:计算每一对字符对答案的贡献。
(从0开始)位置为
是多少个子串的对称位置呢?
个。
如果
,则该对字符对答案的贡献为
.
枚举所有字符对,求和贡献。
import java.io.*;
import java.util.*;
public class Main {
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
public static void main(String[] args) throws Exception {
char[] s = sc.next().toCharArray();
int n = s.length;
long ans = 0;
for(int i = 0; i < n - 1; i ++){
for(int j = i + 1; j < n; j ++){
for(int l = i, r = j; l < r; l ++, r --){
if(s[l] != s[r]) ans ++;
}
}
}
pw.println(ans);
pw.flush();pw.close();
}
}
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import java.io.*;
import java.util.*;
public class Main {
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
public static void main(String[] args) throws Exception {
char[] s = sc.next().toCharArray();
int n = s.length;
int dp[][] = new int[n + 1][n + 1];
long ans = 0;
for(int j = 2; j <= n ; j ++){
for(int i = 1; i <= j - 1; i ++){// left:i right:j
if(s[j - 1] == s[i - 1]) dp[i][j] = dp[i + 1][j - 1];
else dp[i][j] = dp[i + 1][j - 1] + 1;
ans += dp[i][j];
}
}
pw.println(ans);
pw.flush();pw.close();
}
}
------------------------------------------------------------------------------
import java.io.*;
import java.util.*;
public class Main {
static Scanner sc = new Scanner(System.in);
static PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
public static void main(String[] args) throws Exception {
char[] c = sc.next().toCharArray();
int n = c.length;
long ans = 0;
for(int i = 0; i < n - 1; i ++){
for(int j = i + 1; j < n; j ++){
if(c[i] != c[j]) ans += Math.min(i + 1, n - j);
}
}
pw.println(ans);
pw.flush();pw.close();
}
}