找出一个递减序列，假设有两个或两个以上递减序列直接no了，然后对递减序列两端数start。end，然后比較a[start]和a[end+1] 。 a[end] 和a[start-1]

#include<iostream>

#include<stdio.h>

using namespace std;

int a[100005];

int main(){

   // freopen("in.txt","r",stdin);

    int n;

    while(~scanf("%d",&n)){

        a[0]=-1;

        a[n+1]=1000000009;

        for(int i=1;i<=n;i++)

            scanf("%d",&a[i]);

        int start=-1,end=-1,flag=0;

        for(int i=2;i<=n+1;i++){

            if(a[i-1]>a[i] && start==-1){

                start=i-1;

            }

            else if(start!=-1 && end==-1 && a[i-1]<a[i]){

                end=i-1;

                flag=1;

            }

            else if(a[i-1]>a[i] && start!=-1 && end!=-1){

                flag=2;

                break;

            }

        }

   //     cout<<"初始位置："<<start<<" "<<end<<endl;

        if(n==1 || start==end){

            cout<<"yes"<<endl;

            cout<<"1 1"<<endl;

        }

        else if(flag==2){

            cout<<"no"<<endl;

        }

        else if(a[start]<a[end+1] && a[end]>a[start-1]){

            cout<<"yes"<<endl;

            cout<<start<<" "<<end<<endl;

        }

        else

            cout<<"no"<<endl;

    }

}