Super-palindrome HDU - 6264 (改变字符，把每个字符串的奇数子串都变成回文的，一定要善于观察)

题目描述

You are given a string that is consisted of lowercase English alphabet.You are supposed to change it into a super-palindrome string in minimum steps.You can change one character in string to another letter per step.

A string is called a super-palindrome string if all its substrings with an odd length are palindrome strings.That is,for a string s,if its substrings S（i...j ）satisfies j-i+1 is odd then Si+k）=S（j-k） for k = 0,1,...,j-i+1.

输入

The first line contains integer T(1≤T≤100) representing the number of test case.

For each test case,the only line contains a string,which consists of only lowercase letters.It is guaranteed that the length of string satisfies 1≤|s|≤100.

输出

For each test case,print one line with an intger refers to the minimum steps to take.

题意：输入一个t，表示有几组数据，下面输入一个长度不超过 100的字符串，让你改变其中最少的字符，使得只要是长度为奇数的子串都是回文字符串；

思路：这道题，你一定要善于观察，一定要抓住字眼，长度为奇数的子串都是回文字符串；

超级回文串：只要是长度为奇数的子串都是回文字符串；

都是先定义为从下标为1开始的子串

长度为 3 时  要想使回文序列的话 a[1] = a[3]; 长度增长

长度为5 时 要想回文序列 a[1]=a[5] a[2]=a[4],长度为3时，要想使回文 ，a[1]=a[3],那么现在，a[1]=a[5] =a[3] 长度增长

长度为7时  a[1] = a[7],a[2] = a[6],a[3] = a[5], 综合 长度为3和为5知，a[1] = a[3] =a[5] =a[7],a[2] = a[4] = a[6];

可以不是从下标为1的子串开始，可是只要的道理；

综上： 要是是超级回文串，所有的奇数为必须相同，所有的偶数时必须相同

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

#define Max 110
char str[Max];
int main()
{
	int i,j,k,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",str);
		int l = strlen(str);
		int a[30];
		memset(a,0,sizeof(a));
		for(i = 0;i<l;i+=2)
			a[str[i]-'a']++;
			
		int sum = 0;
		int Ma = 0;
		for(i =0;i<26;i++)
			Ma = max(Ma,a[i]);
		sum +=Ma;
		memset(a,0,sizeof(a));
		for(i = 1;i<l;i+=2)
			a[str[i]-'a']++;
		 Ma = 0;
		for(i=0;i<26;i++)
			Ma = max(Ma,a[i]);
		sum +=Ma;
		printf("%d\n",l-sum);
	}
	return 0;
}