Codeforces Round #323 (Div. 2) E - Superior Periodic Subarrays

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> >

using namespace std;

const int N = 2e5 + 10;
const int M = 10 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int gcd(int a, int b) {
    return !b ? a : gcd(b, a % b);
}

int n, mx, a[N << 1], cnt[N << 1], f[N << 1];
bool flag[N << 1];

int main() {
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
        a[i + n] = a[i];
    }
    
    LL ans = 0;
    for(int d = 1; d < n; d++) { //枚举GCD(n,s)
        if(n % d) continue;

        memset(flag, false, sizeof(flag));

        for(int k = 0; k < d; k++) { //枚举起点
            mx = 0;
            for(int i = k; i < 2 * n; i += d) mx = max(mx, a[i]); //找出这些相隔为d的数之间最大的数 
            for(int i = k; i < 2 * n; i += d) flag[i] = (a[i] == mx);//判断这个数是不是一系列数的最大值
        }

        f[0] = flag[0];  //以i结尾的"卓越子序列"最长可能长度 
        for(int i = 1; i < 2 * n; i++) {
            if(flag[i]) f[i] = f[i - 1] + 1;
            else f[i] = 0;
            f[i] = min(f[i],  n - 1);
        }

        cnt[0] = 0; //1-i中有多少个数与n的gcd为枚举的d   
        for(int i = 1; i < n / d; i++) cnt[i] = cnt[i - 1] + (gcd(i, n / d) == 1);
        //要把Gcd(s,n)==d化简成gcd(s/d,n/d)==1，不然会被卡
        
        
        for(int i = n; i < n * 2; i++) ans += cnt[f[i] / d];
    }

    printf("%lld\n", ans);
    return 0;
}
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