hihocoder1831 80 Days

思路：

令p[i] = a[i] - b[i]，p[i + n] = p[i]（i = 1,2,...,n），则需要找出一段长度为n的连续序列使此序列的任一前缀和均大于-c。转化如下：首先求序列p的前缀和sum[i]（i = 1,2,...,2 * n - 1）,然后对i = 1,2,...,n，检查是否sum[i]...sum[i + n - 1]中的每个数均大于或等于sum[i - 1] - c即可。可以使用单调队列预处理n - 1个区间最小值。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int MAXN = ;
 ll a[MAXN * ], q[MAXN * ], minn[MAXN], sum[MAXN * ];
 int main()
 {
     ios::sync_with_stdio(false);
     int T, n;
     ll c;
     cin >> T;
     while (T--)
     {
         memset(q, , sizeof q);
         memset(minn, , sizeof minn);
         memset(sum, , sizeof sum);
         cin >> n >> c;
         ll x;
         for (int i = ; i <= n; i++) cin >> a[i];
         for (int i = ; i <= n; i++)
         {
             cin >> x;
             a[i] -= x;
             a[i + n] = a[i];
         }
         for (int i = ; i <=  * n - ; i++) sum[i] = sum[i - ] + a[i];
         int front = , back = ;
         q[front] = ;
         for (int i = ; i <=  * n - ; i++)
         {
             while (sum[i] <= sum[q[back]] && back >= front)
                 back--;
             q[++back] = i;
             while (q[front] <= i - n && front < back)
                 front++;
             if (i >= n) minn[i - n + ] = sum[q[front]];
         }
         int i = ;
         for ( ; i <= n; i++)
         {
             if (minn[i] >= sum[i - ] - c) break;
         }
         cout << (i == n +  ? - : i) << endl;
     }
     return ;
 }