Educational Codeforces Round 92 (Rated for Div. 2) C - Good String (思惟)

题目传送
题意:
给你一个字符串，如今有俩种操做，一种是字符串全部位都向左平移一个单位，另外一种是向右平移一个单位，如今要使得这个字符串俩种操做后的字符串仍是相等的，问最少改动多少个字符?

思路:
既然要平移后的字符串仍是彻底相等的，那么能够获得:
s[i-1] == s[i+1] (向右平移的和向左平移的)，那么再推一下就能够获得只有俩种字符串是能够的:
1111111 或者 2525252525(这种只能是偶数)
那么因为是只有0到9的数字组成，那么固然也就20种状况，暴力跑一下就能够了

AC代码

#include <bits/stdc++.h>
inline long long read(){char c = getchar();long long x = 0,s = 1;
while(c < '0' || c > '9') {if(c == '-') s = -1;c = getchar();}
while(c >= '0' && c <= '9') {x = x*10 + c -'0';c = getchar();}
return x*s;}
using namespace std;
#define NewNode (TreeNode *)malloc(sizeof(TreeNode))
#define Mem(a,b) memset(a,b,sizeof(a))
#define lowbit(x) (x)&(-x)
const int N = 2e5 + 5;
const long long INFINF = 0x7f7f7f7f7f7f7f;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-7;
const int mod = 1e9 + 7;
const double II = acos(-1);
const double PP = (II*1.0)/(180.00);
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> piil;
signed main()
{
    std::ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    int t;
    cin >> t;
    while(t--)
    {
        string s;
        cin >> s;
        int Min = INF,len = s.size();
        for(int i = 0;i < 10;i++)
        {
            for(int j = 0;j < 10;j++)
            {
                int num = 0,x = 0,y = 0;
                for(int k = 0;k < len;k++)
                {
                    if(!x && s[k]-'0' == i)
                        num++,x = 1;
                    else if(x && !y && s[k]-'0' == j)
                        num++,y = 1;
                    if(x && y)
                        x = 0,y = 0;
                }
                if(i != j && num % 2 != 0) num--;//判断是哪一种状况
                Min = min(Min,len-num);//更新改动的最少次数
            }
        }
        cout << Min << endl;
    }
}