HDU - 6264 Super-palindrome 【思维】

题目

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 substring si···j satisfies j | i + 1 is odd then si+k = sj| k for
k = 0, 1, · · · , j | i + 1.
Input
The first line contains an integer T (1 ≤ T ≤ 100) representing the number of test cases.
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.
Output
For each test case, print one line with an integer refers to the minimum steps to take

输入
3
ncncn
aaaaba
aaaabb
输出
0
1
2

题意：t组输入，每次输入一个字符串，问在这个字符串中想让j-i+1=奇数的所有位置字母相等最少进行几次操作，每次操作可以把把串中的一个字母改变

思路：符合上述位置要求的下标只能有两种序列，1,3，5,7，9…；2,4，6,8，10…
下标从零开始的话就是0，2,4，6,8，10…；1,3，5,7，9…让操作次数相等那我们肯定要不操作在符合位置上某种最多的字母，剩下的都要操作，总长度减去两种序列上两种最多的两种字母的出现次数即为所求

AC code

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<sstream>
#include<queue>
#include<stack>
using namespace std;
char a[120];
int b[250],c[250];
void solve()
{
    int k=strlen(a);
    int max1=0,max2=0;
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    for(int i=0;i<k;i++)//求两种序列上出现次数最多的两种字母
      if(i%2==0)
      {
          b[a[i]]++;
          max1=max(max1,b[a[i]]);
      }
        else
        {
             c[a[i]]++;
          max2=max(max2,c[a[i]]);
        }
        int s=k-max1-max2;//最小操作数
        printf("%d\n",s);
}
int main()
{
    ios::sync_with_stdio(0);
    int t;
    cin>>t;
    while(t--)
    {
        cin>>a;
        solve();
    }
}