标签：mini empty and isp namespace get das exactly hat
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
5 2
1 2 3 4 5
5
5 1
-4 -5 -3 -2 -1
-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
水题，关键是搞清楚题意，还有当k==2时，为什么是max（a[0], a[n-1]）,为什么a[0],和a[n-1]一定是序列里的最小值？？？？
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; int main() { int n,k; int i,j; int ans; int mmax,mmin; int a[100010]; mmax = -1e9-7; mmin = 1e9+7; scanf("%d %d",&n, &k); for(i = 0; i < n; i++) { scanf("%d",&a[i]); } if(k == 1) { for(i = 0; i < n; i++) { mmin = min(a[i], mmin); } printf("%d\n",mmin); } else if(k == 2) { ans = max(a[0], a[n-1]); printf("%d\n",ans); } else if(k >= 3) { for(i = 0; i < n; i++) { mmax = max(a[i], mmax); } printf("%d\n",mmax); } return 0; }
C - Maximum of Maximums of Minimums（数学）
标签：mini empty and isp namespace get das exactly hat
原文地址：http://www.cnblogs.com/ruruozhenhao/p/7744781.html
