Codeforces Round #258 (Div. 2/B)/Codeforces451B_Sort the Array,

Codeforces Round #258 (Div. 2/B)/Codeforces451B_Sort the Array,

Solution report

Http://blog.csdn.net/juncoder/article/details/38102391

For the given array, take the segment in the array for flip and ask if it is incremental and ordered after flip.

Ideas:

You only need to find the area that is originally decreasing, record the largest and smallest values in the area, and the interval position.

Then compare the maximum value with the next element of the interval, and the minimum value with the previous element of the interval.

This is not enough. There may be two or more decline intervals. In this case, pass directly because it can only be flipped once.

#include <iostream>#include <cstdio>#include <cstring>#define inf 999999999999#define LL long long#define swap(x,y,t) ((t)=(x),(x)=(y),(y)=(t))using namespace std;LL num[100010];int main(){    int n,i,j;    while(cin>>n)    {        for(i=1; i<=n; i++)            cin>>num[i];        for(i=1; i<=n; i++)        {            if(num[i+1]<num[i])                break;        }        num[n+1]=inf;        if(i==n+1)        {            printf("1 1\n");            continue;        }        int l,r;        LL maxx=0,minn=inf;        int t=i;        for(; i<=n; i++)        {            if(maxx<num[i])            {                maxx=num[i];                r=i;            }            if(minn>num[i])            {                l=i;                minn=num[i];            }            if(num[i+1]>num[i])                break;        }        int u=i;        int q=0;        for(; i<=n; i++)        {            if(num[i+1]<num[i])            {                q=1;                printf("no\n");                break;            }        }        if(!q)            if(num[u+1]>=maxx&&num[t-1]<minn)            {                printf("yes\n");                if(l<r)                    swap(l,r,q);                printf("%d %d\n",r,l);            }            else printf("no\n");    }    return 0;}

Sort the Arraytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Being a programmer, you like arrays a lot. For your birthday, your friends have given you an arrayAConsistingNDistinct integers.

Unfortunately, the sizeAIs 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 arrayA(In increasing order) by reversing exactly one segmentA? See definitions of segment and reversing in the notes.

Input

The first line of the input contains an integerN(1 digit ≤ DigitNLimit ≤ limit 105)-the size of arrayA.

The second line containsNDistinct space-separated integers:A[1], bytesA[2], middle..., middleA[N] (1 limit ≤ limitA[I] Limit ≤ limit 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

33 2 1

Output

yes1 3

Input

42 1 3 4

Output

yes1 2

Input

43 1 2 4

Output

no

Input

21 2

Output

yes1 1

Note

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

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

Definitions

A segment [L, Bytes,R] Of arrayAIs the sequenceA[L], RoleA[LKeys + keys 1], keys..., keysA[R].

If you have an arrayAOf sizeNAnd you reverse its segment [L, Bytes,R], The array will become:

A[1], bytesA[2], middle..., middleA[LAccept-limit 2], denyA[LExecutor-cores 1], BuffersA[R], RoleA[RAudio-extract 1], audio..., audioA[LLimit + limit 1], limitA[L], RoleA[RLimit + limit 1], limitA[RPipeline + pipeline 2], pipeline..., PipelineA[NExecutor-cores 1], BuffersA[N].