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;
}
}
