救救孩子把，调了一上午了

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 1e5 + 5;
int n, k, a[N], b[N], l[N], r[N], sum[N], res;

stack<int> s;

signed main() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> b[i], sum[i] = sum[i - 1] + b[i];
    a[0] = a[n + 1] = 0;
    for (int i = 1; i <= n; i++) {
        if (s.size() && a[s.top()] >= a[i]) s.pop();
        if (s.size()) l[i] = s.top();
        else l[i] = i - 1;
        s.push(i);
    }
    while (s.size()) s.pop();
    for (int i = n; i; i--) {
        if (s.size() && a[s.top()] >= a[i]) s.pop();
        if (s.size()) r[i] = s.top();
        else r[i] = i + 1;
        s.push(i);
    }
    for (int i = 1; i <= n; i++) {
        int ls = l[i] + 1, rs = r[i] - 1;
        if (i - ls < rs - i) {
            for (int j = ls; j <= i; j++) {
                int lq = i, rq = rs;
                while (lq < rq) {
                    int mid = (lq + rq + 1) >> 1;
                    if (sum[mid] - sum[j - 1] + a[i] <= k) lq = mid;
                    else rq = mid - 1;
                }   
                res += (lq - i + 1);
            } 
        }
        else {
            for (int j = i; j <= rs; j++) {
                int lq = ls, rq = i;
                while (lq < rq) {
                    int mid = (lq + rq) >> 1;
                    if (sum[j] - sum[mid - 1] + a[i] <= k) rq = mid;
                    else lq = mid + 1;
                }
                res += (i - lq + 1);
            }
        }
    }
    cout << res << endl;
}

做法是从 min 入手，假设 $a_i=\min\{a_l,a_{l+1},....,a_r\}$，左边右边找到 $l,r$ 的边界，再来二分