CF1237D Balanced Playlist

思路：
假设从第i首歌开始听，结束位置为j，那么从第i+1首歌开始听，结束位置一定不早于j。可以用反证法证明。想到这一点，就不难解决了。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 const int INF = 0x3f3f3f3f;
 const int N = ;
 int a[N], st[N][];
 int log2(int x)
 {
     int res = -;
     while (x) { x >>= ; res++; }
     return res;
 }
 void init(int n)
 {
     for (int i = ; i < n; i++) st[i][] = a[i];
     for (int j = ; ( << j) < n; j++)
     {
         for (int i = ; i + ( << j) -  < n; i++)
         {
             st[i][j] = max(st[i][j - ], st[i + ( << j - )][j - ]);
         }
     }
 }
 int get_max(int l, int r)
 {
     if (l > r) return -INF;
     int p = log2(r - l + );
     return max(st[l][p], st[r - ( << p) + ][p]);
 }
 int main()
 {
     int n;
     while (cin >> n)
     {
         int minn = INF, maxn = -INF;
         for (int i = ; i < n; i++)
         {
             cin >> a[i]; a[i + n] = a[i +  * n] = a[i];
             minn = min(a[i], minn);
             maxn = max(a[i], maxn);
         }
         init( * n);
         vector<int> res(n, -);
         if (maxn <= minn * )
         {
             for (auto it: res) cout << it << " ";
             cout << endl; continue;
         }
         int cur = ;
         for (int i = ; i < n; i++)
         {
             cur = max(cur, i + );
             int maxn = get_max(i, cur);
             while (cur <  * n && a[cur] *  >= maxn)
             {
                 maxn = max(maxn, a[cur]); cur++;
             }
             res[i] = cur - i;
         }
         for (auto it: res) cout << it << " ";
         cout << endl;
     }
     return ;
 }