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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 0x3f3f3f3fconst 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); }}
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