CSU-ACM暑假集训基础组训练赛(2) B - Problem B

B - Problem B
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice CodeForces
451B

Description

Being a programmer, you like arrays a lot. For your birthday, your friends have given you an array a consisting of ndistinct 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 Input

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

Hint

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].

模拟+数组排序

代码如下：

#include <iostream>

#include<cstdio>

#include <algorithm>

using namespace std;

int const maxn=100000+10;

struct node

{

int val,id;

bool operator <(const node &rhs) const

{

return val<rhs.val;

}

}nod[maxn];

int main()

{

int n,i,r,l;

bool flag=1;

scanf("%d",&n);

for(i=1;i<=n;i++)

{

scanf("%d",&nod[i].val);

nod[i].id=i;

}

//nod[n+1].id=-1;

sort(nod+1,nod+n+1);

nod[n+1].id=-1;

l=-1;

for(i=1;i<=n;i++)

if(nod[i].id!=i)

{

l=i;break;

}

if(l==-1)

{

printf("yes\n1 1\n");

return 0;

}

for(i=l;i<=n;i++)

if(nod[i].id-nod[i+1].id!=1)

{

r=i+1;break;

}

// if(i==n+1){printf("yes\n%d %d\n",l,r-1);return 0;}//加上这句虽然没错，但是多余了

for(i=r;i<=n;i++)

if(nod[i].id!=i)

{

//printf("no\n");return 0;

flag=false;//或者flag=0

break;

}

if(flag) printf("yes\n%d %d\n",l,r-1);

else printf("no\n");

return 0;

}

总结：

主要是熟悉一下结构体中重载运算符<用于sort函数的方法。重视基础，方法得当，事半功倍。