寒假训练，2.25，J-Palindrome Names （回文-优快云博客

本文介绍了一种算法，用于计算将任意名字最少修改几次可变为回文串的方法。通过对比字符并计数非匹配项，该算法能高效确定所需更改次数。

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题目：

Anna and Bob are having a baby. They both enjoy the advantage of having palindrome names, meaning that their names are spelled the same way forwards and backwards. Wanting to be good parents, they decide to give their child a palindrome name too. The only problem is that they aren’t sure if the one they picked is a palindrome. If it turns out it isn’t a palindrome, they want to change it to a palindrome using as few changes as possible. The allowed changes are:

Change one letter of the name.
Add a letter to the end of the name.

Help Bob and Anna find out how many changes they need to make to the name to make it a palindrome.

Input

Input is the name they have chosen.

Output

Output the number of changes they need to make.

Limits

The length of the name is at least 1
The name consists of only lowercase letters a–z.

#include<cstdio>
#include<string.h>
#include<algorithm>
using namespace std;
char str[102];
int num[102];
int cntt(int i,int j){
    int cnt=0;
    while(i<j){
        if(str[i]!=str[j-1]){cnt++;}
        i++;j--;
    }
    return cnt;
}
int main(){
    while(~scanf("%s",str)){
        int minn;
        int len=strlen(str);
        for(int i=0; i<len/2; i++){
            num[i]=i+cntt(i,len);
		}
        minn=num[0];
        for(int i=1;i<len/2;i++){
            minn=min(minn,num[i]);
        }
        printf("%d\n",minn);
    }
}

　　我到现在没想通什么原理，但真的好方便啊

转载于:https://www.cnblogs.com/ineedyou/p/8471200.html