D. Max GEQ Sum

我们考虑暴力枚举a[i]为最大值 通过单调栈可以求出a[i]左边右边第一个大于a[i]的 然后通过ST表查询前缀和数组（i，R[i]-1）的最大值 （L[i]+1,i)的最小值得到我们需要的区间和最大值 check即可
注意我们这里因为是前缀和 query_max(i, R[i] - 1) - query_min(L[i], i - 1) L这边要-1 所以就会有0这个点 我们ST表的范围就要更改为[0,n]

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
const int M = 998244353;
const int mod = 1000000007;
#define int long long
#define endl '\n'
#define Endl '\n'
#define YES cout<<"YES"<<endl;
#define NO cout<<"NO"<<endl;
#define _ 0
#define inf 0x3f3f3f3f3f3f3f3f
#define fast ios::sync_with_stdio(false);cin.tie(nullptr);
int a[N],f1[N][30],f2[N][30],n,L[N],R[N],pre[N];
void init() {
    for (int len = 0; len < 30; len++) {
        for (int i = 0; i + (1 << len) - 1 <= n; i++) {
            if (!len)f1[i][len] = pre[i];
            else f1[i][len] = max(f1[i][len - 1], f1[i + (1 << (len - 1))][len - 1]);
        }   //i+1<<len-1的话是包括了i 不用加1
    }
    for (int len = 0; len < 30; len++) {
        for (int i = 0; i + (1 << len) - 1 <= n; i++) {
            if (!len)f2[i][len] = pre[i];
            else f2[i][len] = min(f2[i][len - 1], f2[i + (1 << (len - 1))][len - 1]);
        }   //i+1<<len-1的话是包括了i 不用加1
    }
}
int query_max(int l,int r) {
    int len = r - l + 1;
    int k = log(len) / log(2);
    return max(f1[l][k], f1[r - (1 << k) + 1][k]); //而r-1<<k是没有包含到r所以要+1
}
int query_min(int l,int r) {
    int len = r - l + 1;
    int k = log(len) / log(2);
    return min(f2[l][k], f2[r - (1 << k) + 1][k]); //而r-1<<k是没有包含到r所以要+1
}
void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++)cin >> a[i], pre[i] = pre[i - 1] + a[i];
    stack<int> stk;
    a[n + 1] = inf;
    for (int i = 1; i <= n + 1; i++) {
        while (stk.size() && a[stk.top()] < a[i]) {
            R[stk.top()] = i;
            stk.pop();
        }
        stk.push(i);
    }
    while (stk.size())stk.pop();
    a[0] = inf;
    for (int i = n; i >= 0; i--) {
        while (stk.size() && a[stk.top()] < a[i]) {
            L[stk.top()] = i;
            stk.pop();
        }
        stk.push(i);
    }
    for (int i = 1; i <= n; i++) {
        if (a[i] < (query_max(i, R[i] - 1) - query_min(L[i], i - 1))) {
            NO
            return;
        }
    }
    YES
}
signed main(){
    fast
    int T;cin>>T;
    while(T--) {
        solve();
    }
    return ~~(0^_^0);
}