The original question link
idea: I
got a red question when I woke up, and when I saw that I didn't open ll, I was careless.
Just construct a sequence of 0, 1, 2..., if there are fewer, see if the previous ones can be filled.
Code:
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;typedef unsigned long long ull;
typedef pair<ll,ll>PLL;typedef pair<int,int>PII;typedef pair<double,double>PDD;
#define I_int ll
inline ll read(){
ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){
if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=x*10+ch-'0';ch=getchar();}return x*f;}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a,ll b,ll p){
ll res=1;while(b){
if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;}
const int maxn=1e6+7;
ll a[maxn],n;
void solve(){
n=read;
rep(i,1,n) a[i]=read;
ll las=-1,sum=0;
rep(i,1,n){
las++;
if(a[i]>=las){
sum+=a[i]-las;
}
else{
int t=las-a[i];
if(sum>=t) sum-=t;
else{
puts("NO");
return ;
}
}
}
puts("YES");
}
int main(){
int T=read;
while(T--) solve();
return 0;
}