思路：求最大最小表示的位置，直接模板，求次数的话，先用KMP求出循环节的大小，再判断一下，最大最小表示出现次数相同，具体看代码注释。

AC code:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
const int maxn = 1e6 + 10; 
string str;
int len = 0;
int nex[maxn];
void make_nex(){//KMP的nex数组 
    int j = 0,k = -1;
    nex[0] = -1;
    while(j <= len){
        if(k == -1 || str[k] == str[j]){
            if(str[k+1] == str[j+1]){
                nex[j+1] = nex[k+1];
                ++j,++k;
            }else{
                nex[j+1] = k+1;
                ++j,++k;
            }
        }else{
            k = nex[k];
        }
    }
}
int min_rep(){
    int i = 1,j = 0,k = 0;

    while(j < len && i < len){
        k = 0;          //之前这个忘记写了 
        while(str[(i+k) % len] == str[(j+k) % len] && k < len){
            ++k;
        }
        if(k == len){
            return min(i,j) + 1;
        }
        if(str[(i+k)%len] > str[(j+k)%len]){
            i = max(i+k+1, j+1);
        }else{
            j = max(j + k + 1, i + 1);
        }
    }
    return min(i,j) + 1;

}
int max_rep(){
    int i = 1,j = 0,k = 0;

    while(j < len && i < len){
        k = 0; 
        while(str[(i+k) % len] == str[(j+k) % len] && k < len){
            ++k;
        }
        if(k == len){
            return min(i,j) + 1;
        }
        if(str[(i+k)%len] < str[(j+k)%len]){
            i = max(i+k+1, j+1) ;
        }else{
            j = max(j + k + 1, i + 1);
        }
    }
    return min(i,j) + 1;


}
int main(){
    while(cin >> str){
        len = str.size(); 
        memset(nex,0,sizeof(nex));
        make_nex();         //之前忘记，调用函数了，也忘记了初始化 
        int min_pos = min_rep();
        int  max_pos = max_rep();
        int min_times = 0,max_times = 0;
        int r = len - nex[len];//求循环节的大小 
        //cout << "enm " << r << endl; 
        if(len % r == 0){//别忘记判断，要的是纯净的循环，全有循环节组成，不含杂质 
            min_times = max_times = len / r;
        }else{
            min_times = max_times = 1;
        }
        //cout << min_rep() << endl;
        cout << min_pos << " " << min_times << " " << max_pos << " " << max_times << endl;
    }



    return 0;
}