今日SGU 5.19

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+7;
const int maxn = 5e5+6;
const double eps = 1e-6;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while (ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
//这么考虑，一个串（长度为n）拥有长度为k的子串种类最多有n-k+1种，那么题目里的n最多为5e5，
//那么2的19次方就已经超过了5e5了，那么我们找的子串长度最多为19
//然后你在预处理一下就行了
char s[maxn];
bool f[(1<<19)+6][20] = {0};
int main(){
    int n;
    scanf("%d%s",&n,s);
    rep(i,0,n){
        int tmp = 0;
//        de(i)
        rep(j,1,20){
            if(i + j > n) break;
            tmp = tmp * 2 + s[i+j-1] - ‘a‘;
//            dd(j)de(tmp)
            f[tmp][j] = true;
        }
//        cout<<endl;
    }
    rep(ans,1,20){
//        de(ans)
        rep(i,0,1<<ans){
//            d0e(i)
            if(!f[i][ans]){
                printf("%d\n",ans);
                repd(k,ans-1,0) {
                    if(i&(1<<k)) printf("b");
                    else printf("a");
                }
                return 0;
            }
        }
    }
    return 0;
}