AtCoder AGC030B Tree Burning

Topic Link

https://atcoder.jp/contests/agc030/tasks/agc030_b

answer

Good question details. .
First, assume that the first step right away, you can find a certain number of times when the turn when the path is only
then can enumerate the value
and annoying to sub-parity after the enumeration of discussion. .
Finally, the ability to turn over and do it again the first step processing left to go on the trip
time complexity \ (O (n) \)

Code

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define llong long long
using namespace std;

void read(int &x)
{
    int f=1;x=0;char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
    x*=f;
}

const int N = 2e5;
llong a[N+3];
llong s[N+3];
int n; llong m,ans;

void solve()
{
    s[0] = 0ll; for(int i=1; i<=n; i++) s[i] = s[i-1]+a[i];
    for(int i=1; i<=n; i++)
    {
        llong tmp;
        if(i&1)
        {
            int x = (i-1)>>1;
            tmp = (m*x-(s[n]-s[n-x]))*2+a[n-x]+(s[n-x-1]-s[n-x-x-1])*2;
        }
        else
        {
            int x = i>>1;
            tmp = (m*(x-1)-(s[n]-s[n-x+1]))*2+(m-a[n-x+1])+(s[n-x]-s[n-x-x])*2;
        }
        ans = max(ans,tmp);
    }
}

int main()
{
    scanf("%lld%d",&m,&n);
    for(int i=1; i<=n; i++) scanf("%lld",&a[i]);
    solve();
    for(int i=1; i<=n; i++) a[i] = m-a[i];
    reverse(a+1,a+n+1);
    solve();
    printf("%lld\n",ans);
    return 0;
}