Domino for Young

D. Domino for Young
You are given a Young diagram.

Given diagram is a histogram with n columns of lengths (a_1,a_2,…,a_n (a_1≥a_2≥…≥a_n≥1).)

Young diagram for 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 1×2 or 2×1 rectangle.

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

The next line of input contains n integers (a_1,a_2,…,a_n (1≤a_i≤300000,a_i≥a_{i+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
inputCopy
5
3 2 2 2 1
outputCopy
4
Note
Some of the possible solutions for the example:

如上图, 考虑二分图匹配, 红点和旁边白点连边, 但是(n<9e^{10}), 跑不动
手动模拟匹配

假设红点数小于等于白点数且当前有红点h未匹配(显然还有白点b未匹配) 那么,

1. 如果h旁边有白点为匹配, 匹配数+1
2. 如果h旁边所有白点都匹配了, 则h->b的路径必然是
   h --->B->H->B ....->H->B->H------->b路径,那么,这就是一条可增广路,匹配数+1

所以, 只要有足够多的的红点, 所有白点都能匹配
红白点可以互换.则求红白点数的最小值

#include<bits/stdc++.h>
using namespace std;
char buf[1<<20],*_=buf,*__=buf;
#define gc() (_==__&&(__=(_=buf)+fread(buf,1,1<<20,stdin),_==__)?EOF:*_++)
#define TT template<class T>inline
TT bool read(T &x){
    x=0;char c=gc();bool f=0;
    while(c<48||c>57){if(c==EOF)return 0;f^=(c=='-'),c=gc();}
    while(47<c&&c<58)x=(x<<3)+(x<<1)+(c^48),c=gc();
    if(f)x=-x;return 1;
}
TT bool read(T&a,T&b){return read(a)&&read(b);}
TT bool read(T&a,T&b,T&c){return read(a)&&read(b)&&read(c);}
typedef long long ll;
const ll MAXN=1e1+8,mod=1e9+7,inf=0x3f3f3f3f;
int main() {
    ll n,x,f=0,res1=0,res2=0;//红白点数量.
    read(n);
    while(n--){
        read(x);
        if(f){
            res1+=(x+1)>>1;
            res2+=x>>1;
        }
        else{
            res1+=x>>1;
            res2+=(x+1)>>1;
        }
        f^=1;
    }
    cout<<min(res1,res2);
    return 0;
}