CodeForces - 870B Maximum of Maximums of Minimums

Maximum of Maximums of Minimums 

You are given an array a1, a2, ..., an consisting of n integers, and an integer k. You have to split the array into exactly k non-empty subsegments. You'll then compute the minimum integer on each subsegment, and take the maximum integer over the k obtained minimums. What is the maximum possible integer you can get?
Definitions of subsegment and array splitting are given in notes.
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤  105) — the size of the array a and the number of subsegments you have to split the array to.
The second line contains n integers a1,  a2,  ...,  an ( - 109  ≤  ai ≤  109).
Output
Print single integer — the maximum possible integer you can get if you split the array into k non-empty subsegments and take maximum of minimums on the subsegments.
Example
Input
5 2
1 2 3 4 5
Output
5
Input
5 1
-4 -5 -3 -2 -1
Output
-5
Note
A subsegment [l,  r] (l ≤ r) of array a is the sequence al,  al + 1,  ...,  ar.
Splitting of array a of n elements into k subsegments [l1, r1], [l2, r2], ..., [lk, rk] (l1 = 1, rk = n, li = ri - 1 + 1 for all i > 1) is k sequences (al1, ..., ar1), ..., (alk, ..., ark).
In the first example you should split the array into subsegments [1, 4] and [5, 5] that results in sequences (1, 2, 3, 4) and (5). The minimums are min(1, 2, 3, 4) = 1 and min(5) = 5. The resulting maximum is max(1, 5) = 5. It is obvious that you can't reach greater result.
In the second example the only option you have is to split the array into one subsegment [1, 5], that results in one sequence ( - 4,  - 5,  - 3,  - 2,  - 1). The only minimum is min( - 4,  - 5,  - 3,  - 2,  - 1) =  - 5. The resulting maximum is  - 5.

题意：把一组数分为k组，顺序不能改变，找到各个组的最小值中的最大值。

思路：按k的值可以分为几种情况，很容易想到，当k=2的时候要找遍历一遍找到最大的那个最小值。

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#define inf 0x3f3f3f3f
const int MAX=1000000;
using namespace std;
long long a[MAX];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        long long x=-inf,y=inf;
        int m=-1;
        memset(a,0,sizeof(a));
        for(int i=1; i<=n; i++)
        {
            cin>>a[i];
            if(a[i]>=x)
            {
                x=a[i];
                m=i;
            }
            if(a[i]<y)
            {
                y=a[i];
            }
        }
        if(k>=3||n==1)
        {
            printf("%lld\n",x);
        }
        else if(k==2)
        {
            if(m==1||m==n)
                printf("%lld\n",x);
            else
            {
                long long ans=inf;
                long long ans2=-inf;
                for(int i=1; i<=n; i++)
                {
                    ans=min(ans,a[i]);
                    ans2=max(ans,ans2);
                }
                long long ans1=inf;
                for(int i=n; i>=1; i--)
                {
                    ans1=min(ans1,a[i]);
                    ans2=max(ans2,ans1);
                }
                printf("%lld\n",ans2);
            }
        }
        else if(k==1)
            printf("%lld\n",y);
    }
}