codeforces 1389C Good String 枚举

易证明，最后的字符串形式一定是 abab(ldots)ab 或 aaa(ldots)aaa 的形式
所以枚举 ab
注意 0a 的形式

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 200010;

int n; 
char s[maxn];

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }

int main(){
	n = read();
	
	for(int i=1;i<=n;i++){
		scanf("%s",s+1);
		int len = strlen(s+1);
		int mi = 1000000007;
		for(int k=10;k<=99;k++){
			int t = 0,f = 0;
			int a = k/10,b = k%10;
			for(int j=1;j<=len;j++){
				if(!f){
					if(s[j]-'0' == a) f = !f;
				}else{
					if(s[j]-'0' == b){
						f = !f;
						++t;
					}
				}
			}
//			printf("t:%d
",t);
			mi = min(mi,len-2 * t);
		}
		
		for(int k=0;k<=9;k++){
			int t = 0,f = 0;
			int a = 0,b = k;
			for(int j=1;j<=len;j++){
				if(!f){
					if(s[j]-'0' == a) f = !f;
				}else{
					if(s[j]-'0' == b){
						f = !f;
						++t;
					}
				}
			}
			mi = min(mi,len-2 * t);
		}
		
		int cnt;
		for(int k=0;k<=9;k++){
			cnt = 0;
			for(int j=1;j<=len;j++){
				if(s[j]-'0' == k) ++cnt;
			}
			mi = min(mi,len-cnt);
		}
		
		printf("%d
",mi);
	}
	
	return 0;
}