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

E - Superior Periodic Subarrays

好难的一题啊。。。

这个博客讲的很好，搬运一下。

https://blog.csdn.net/thy_asdf/article/details/49406133

#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 + ;
const int M =  + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + ;
const double eps = 1e-;
 
int gcd(int a, int b) {
    return !b ? a : gcd(b, a % b);
}
 
int n, mx, a[N << ], cnt[N << ], f[N << ];
bool flag[N << ];
 
int main() {
    scanf("%d", &n);
    for(int i = ; i < n; i++) {
        scanf("%d", &a[i]);
        a[i + n] = a[i];
    }
 
    LL ans = ;
    for(int d = ; d < n; d++) { //枚举GCD(n,s)
        if(n % d) continue;
 
        memset(flag, false, sizeof(flag));
 
        for(int k = ; k < d; k++) { //枚举起点
            mx = ;
            for(int i = k; i <  * 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 <  * 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[] = flag[];  //以i结尾的"卓越子序列"最长可能长度
        for(int i = ; i <  * n; i++) {
            if(flag[i]) f[i] = f[i - ] + ;
            else f[i] = ;
            f[i] = min(f[i],  n - );
        }
 
        cnt[] = ; //1-i中有多少个数与n的gcd为枚举的d
        for(int i = ; i < n / d; i++) cnt[i] = cnt[i - ] + (gcd(i, n / d) == );
        //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 * ; i++) ans += cnt[f[i] / d];
    }
 
    printf("%lld\n", ans);
    return ;
}
/*
*/