codeforces D. Domino for Young (half the match - black and white coloring) Comments

Codeforces Round #609 (Div. 2) D. Domino for Young

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a Young diagram.

Given diagram is a histogram with nn columns of lengths a1,a2,…,ana1,a2,…,an (a1≥a2≥…≥an≥1a1≥a2≥…≥an≥1).

Young diagram for a=[3,2,2,2,1]a=[3,2,2,2,1].

Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1×21×2 or 2×12×1rectangle.

Input

The first line of input contain one integer nn (1≤n≤3000001≤n≤300000): the number of columns in the given histogram.

The next line of input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤300000,ai≥ai+11≤ai≤300000,ai≥ai+1): the lengths of columns.

Output

Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.

Example

input

5
3 2 2 2 1

　　

output

4

　　

Note

Some of the possible solutions for the example:

Meaning of the questions: to give you a chart like the one above of young, before a height greater than equal to the height of a post, asked 2X1 matrix can lay down how many 1X2, or?

This analysis can be seen bipartite matching, because the adjacent two points to match into a matrix, then we build map? There are 5 * 5 th power of two sides * 3 * 3 * 10 10 (not mistaken analysis), Hungary match is O (n * m). I do not think, impossible.

So how can simulate the correct answer? We use black and white staining, direct black and white coloring, color dye out of the fewest matches is a good matrix. For more details, please see the comments:

#include <bits / STDC ++ H.> 
the using namespace STD; 
#define LL Long Long int 
LL A [300005]; 
LL C [2]; 
int main (void) 
{ 
	int n-; 
	Scanf ( "% D", & n-); 
	for (int I =. 1; I <= n-; I ++) { 
		Scanf ( "% LLD", & A [I]); 
		// left to right, the first column above each dyeing begin 
		// dyed white 
		c [0] + = a [i] / 2 ; // white 
		// then dyed black 
		c [1] + = a [ i] / 2; // black 
		 
		if (a [i]% 2 == 1) // this column cardinality two, there must be a point of a polychromatic color 
		// number of columns this point consider only what color dye will do 
		c [i% 2] ++; // dyed white on the provisions of the odd columns, even column dyed black 
		// is because if the line did not have a stain, this line also did not have a stain 
		// order to allow more adjacent matching black and white dots, then only make a front and a rear dyed a different color 
		// printf ( "0: % D. 1:% D \ n-", C [0], C [. 1]); 
	} 
	LL COUNT = min (C [0], C [. 1]); 
	the printf ("% LLD \ n-", COUNT); 
	return 0; 
}

　　Reference to Part: https: //www.cnblogs.com/YDDDD/p/12082619.html