Sequence segments luoguP1182] [Section II

 

Title Description

For a given length is a positive integer number N of columns Ai , want now divided into M ( M ≤ N segment), and each successive requirements, and the maximum and minimum of each segment.

About maximum minimum:

For example a number of columns 42451 to be divided into three stages

Segment which is as follows:

[4 2][4 5][1]

And for the first segment 6, the first two segments and is 9 , the 3 segments and is 1 , and a maximum of 9 .

Segment which is as follows:

[4][2 4][5 1]

And for the first section 4 , the second section is 2 and 6 , the 3 segments and of 6 and a maximum of 6.

And in any case the segment, the maximum value of not less than 6

To get the number of columns can be 42,451 to be divided into three segments, each of the maximum and minimum of 6.

Input Format

The first row 1 comprises two positive integers N, M.

The first row contains 2 N spaces separated by non-negative integer A_i , as meaning the subject.

Output Format

A positive integer, that is how much each segment and maximum minimum.

Sample input and output

Input # 1copy

5 3
4 2 4 5 1

Output # 1copy

6

Description / Tips

For 2 0 % of the data, there are n ≤ 10 ;

For . 4 0 % of the data, there N≤1000 ;

For . 1 0 0 % of the data, there N≤100000, M ≦ N, A_i N ≤ . 1 0 0 0 0 0 and no more than 10 ^ 9 .

Ideas:

　　This question is visible, so that the maximum value of the minimum requirements of each segment, the answer is clear dichotomy, a dichotomy we val, the left point to the maximum value of the sequence, and the right end point for the sequence,

　　Every two minutes, if only two points satisfies into a limited number of times, r = min-1, or r = mid + 1;

　　Scanf without spaces note in parentheses, otherwise no output, and,

　　Read faster in use is to use a semicolon between the number and the number of read, you can not use commas, because C ++ to perform content after the comma, easy pan.

Code:

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
int l,r,n,m,a[100100],yilin,sum,mid;
bool judge(int x)
{
	yilin=0,sum=0;
	for (int i=1;i<=n;i++)
	{
		if(sum+a[i]<=x) sum+=a[i];
		else sum=a[i] , yilin++;
	}
	return (yilin >= m);
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		r+=a[i];
		l=max(l , a[i] );
	}
	while(l<=r)
	{
		 mid=(l+r) / 2;
		if(judge(mid))l = mid + 1;
		else r = mid-1;
	}
	printf("%d\n",l);
	return 0;
}