CF451B Sort the Array 水题

Codeforces Round #258 (Div. 2)

Sort the Array

 

B. Sort the Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Being a programmer, you like arrays a lot. For your birthday, your friends have given you an array a consisting of n distinct integers.

Unfortunately, the size of a is too small. You want a bigger array! Your friends agree to give you a bigger array, but only if you are able to answer the following question correctly: is it possible to sort the array a (in increasing order) by reversing exactly one segment of a? See definitions of segment and reversing in the notes.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 10⁵) — the size of array a.

The second line contains n distinct space-separated integers: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 10⁹).

Output

Print "yes" or "no" (without quotes), depending on the answer.

If your answer is "yes", then also print two space-separated integers denoting start and end (start must not be greater than end) indices of the segment to be reversed. If there are multiple ways of selecting these indices, print any of them.

Sample test(s)

Input

3
3 2 1

Output

yes
1 3

Input

4
2 1 3 4

Output

yes
1 2

Input

4
3 1 2 4

Output

no

Input

2
1 2

Output

yes
1 1

Note

Sample 1. You can reverse the entire array to get [1, 2, 3], which is sorted.

Sample 3. No segment can be reversed such that the array will be sorted.

Definitions

A segment [l, r] of array a is the sequence a[l], a[l + 1], ..., a[r].

If you have an array a of size n and you reverse its segment [l, r], the array will become:

a[1], a[2], ..., a[l - 2], a[l - 1], a[r], a[r - 1], ..., a[l + 1], a[l], a[r + 1], a[r + 2], ..., a[n - 1], a[n].

 

题意：给出一个数字序列，选其中一连续序列反转后能绝对升序，则输出yes并输出反转区间的开头和结尾下标，否则输出no。

题解：水题，从头扫到尾，a[i]<=a[i-1]则标记i-1为反转串的开头，开始检测反转串，当a[i]>a[i-1]时标记a[i-1]为反转串的结尾，然后反转反转串，观察是否绝对升序。（实际上有相等的话肯定没法绝对升序，不过也不需要特殊判断，少打点字吧）

其实有很多方法，可能有复杂度更低的方法判断，不过CF要的是又快又稳，这种水题直接打最直白的方法比较好。

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cstdio>
#include<cmath>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<set>
#include<stack>
#include<queue>
using namespace std;
#define ll long long
#define usint unsigned int
#define mz(array) memset(array, 0, sizeof(array))
#define minf(array) memset(array, 0x3f, sizeof(array))
#define REP(i,n) for(i=0;i<(n);i++)
#define FOR(i,x,n) for(i=(x);i<=(n);i++)
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define WN(x) printf("%d\n",x);
#define RE  freopen("D.in","r",stdin)
#define WE  freopen("1biao.out","w",stdout)

int a[111111];

int main(){
    int n,m=0;
    int i,l=0,r=0;
    scanf("%d",&n);
    scanf("%d",&a[0]);
    for(i=1;i<n;i++){
        scanf("%d",&a[i]);
        if(m==0){
            if(a[i]<=a[i-1]){
                l=i-1;
                m=1;
            }
        }else if(m==1){
            if(a[i]>a[i-1]){
                r=i-1;
                m++;
            }
        }
    }
    if(m==1)r=n-1;
    //cout<<l<<','<<r<<endl;
    for(i=0;i<(r-l+1)/2;i++)
        swap(a[l+i],a[r-i]);
    int ans=1;
//    for(i=0;i<n;i++)
//        cout<<a[i]<<',';
    for(i=1;i<n;i++)
        if(a[i]<=a[i-1]){ans=0;break;}
    if(ans==0)puts("no");
    else printf("yes\n%d %d\n",l+1,r+1);
    return 0;
}