根据题意可知只要存在前后差大于等于2的两个元素即可
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #include<string> #include<stack> #include<queue> #include<vector> #include<cstdlib> //#include<windows.h> #define fi first #define se second #define rep(i,a,b) for(int i=a;i<=b;i++) #define per(i,a,b) for(int i=a;i>=b;i--) #define mes(a,b) memset(a,b,sizeof(a)) using namespace std; typedef long long ll; const int N = 1e6+10; const int INF = 1e9+5; const int inf = - INF; const int mod = 1e9+7; const double pi = acos(-1.0); int a[N]; int main(){ std::ios::sync_with_stdio(false); int T; cin>>T; while(T--){ int n; int flag=0; cin>>n; rep(i,1,n) cin>>a[i]; rep(i,1,n-1) if(abs(a[i+1]-a[i])>=2){ cout<<"YES"<<endl; cout<<i<<' '<<i+1<<endl; flag=1; break; } if(!flag) cout<<"NO"<<endl; } //system("pause"); return 0; }