可以二分b[1]给a[1]的值，看了standing学了种O(n)的办法，每次求a[i]-b[i-1]，若大于0，则累加到下一次(a[i]只能有b[i]和b[i+1])组成，然后判断这个值和b[i]的大小
#include <bits/stdc++.h> using namespace std; typedef long long ll; int main() { ios::sync_with_stdio(0); cin.tie(0); int t; cin >> t; while (t--) { int n; cin >> n; vector<ll> a(n), b(n); ll balance = 0; for (ll& v : a) cin >> v, balance += v; for (ll& v : b) cin >> v, balance -= v; ll worst = 0, ok = 1; for (int i = 1; i <2 * n; i++) { worst = max(0ll, worst) + a[i % n] - b[(i - 1) % n]; if (worst > b[i % n]) ok = 0; } cout << (balance > 0 || ok == 0 ? "NO" : "YES") << endl; } }
