1 solutions

  • 0
    @ 2025-11-5 18:06:07

    C++ :

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <climits>
    #include <cstring>
    #include <string>
    #include <set>
    #include <map>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <list>
    #include <bitset>
    #define rep(i,m,n) for(i=m;i<=n;i++)
    #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
    #define vi vector<int>
    #define pii pair<int,int>
    #define mod 1000000007
    #define inf 0x3f3f3f3f
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define ll long long
    #define pi acos(-1.0)
    const int maxn=1e6+10;
    const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
    using namespace std;
    ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
    int n,m,len1,len,coco[maxn],all;
    char a[maxn],b[maxn];
    bool check(int now)
    {
     
        for(int i=len1;i<now;i++)b[i]=a[i];
        len1=now;
        b[now]=0;//printf("%d %s\n",now,b);
        while(strstr(a+now,b)-(a+now)==0)
            now+=len1;
        int j=0;
        for(int i=now;i<len;i++)
            if(b[j++]!=a[i])return false;
        return true;
    }
    int main()
    {
        int i,j,k,t;
        while(~scanf("%s",a))
        {
            len=strlen(a);
            all=0;len1=0;
            rep(i,0,len-1){if(a[i]>='A'&&a[i]<='Z')a[i]=a[i]-'A'+'a';if(a[i]==a[0]&&i!=0)coco[all++]=i;}
            rep(i,0,all-1)
            {
                if(check(coco[i]))
                {
                    printf("%d\n%s\n",len1,b);
                    break;
                }
            }
            if(i==all)printf("%d\n%s\n",len,a);
        }
        //system ("pause");
        return 0;
    }
    
    • 1

    Information

    ID
    18780
    Time
    1000ms
    Memory
    64MiB
    Difficulty
    (None)
    Tags
    # Submissions
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