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 = 1000005;
ll a[MAXN * 2], q[MAXN * 2], minn[MAXN], sum[MAXN * 2];
int main()
{
    ios::sync_with_stdio(false);
    int T, n;
    ll c;
    cin >> T;
    while (T--)
    {
        memset(q, 0, sizeof q);
        memset(minn, 0, sizeof minn);
        memset(sum, 0, sizeof sum);
        cin >> n >> c;
        ll x;
        for (int i = 1; i <= n; i++) cin >> a[i];
        for (int i = 1; i <= n; i++)
        {
            cin >> x;
            a[i] -= x;
            a[i + n] = a[i];
        }
        for (int i = 1; i <= 2 * n - 1; i++) sum[i] = sum[i - 1] + a[i];
        int front = 1, back = 1;
        q[front] = 1;
        for (int i = 1; i <= 2 * n - 1; 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 + 1] = sum[q[front]];
        }
        int i = 1;
        for ( ; i <= n; i++)
        {
            if (minn[i] >= sum[i - 1] - c) break;
        }
        cout << (i == n + 1 ? -1 : i) << endl;
    }
    return 0;
}