Noi.ac#30 题解

题目链接

(Solution:)

看着题目的约束条件，可以看出是一道二分答案的题目(话说暴力有(90pts?))
我们考虑(a,b)的单调性，我们所取的糖果必然是(a)或(b)的后缀和(不过为什么我用了前缀和来实现)
所以我们分别枚举(a,b)的后缀，对另一个数组二分出愉悦度大于当前枚举数组的最小值，取(max)即可

(Code:)

#include<bits/stdc++.h>
using namespace std;
namespace my_std
{
	typedef long long ll;
	#define fr(i,x,y) for(ll i=(x);i<=(y);i++)
	inline ll read()
	{
		ll sum=0,f=1;
		char ch=0;
		while(!isdigit(ch))
		{
			if(ch=='-') f=-1;
			ch=getchar();
		}
		while(isdigit(ch))
		{
			sum=(sum<<1)+(sum<<3)+(ch^48);
			ch=getchar();
		}
		return sum*f;
	}
	inline void write(ll x)
	{
	    if(x<0)
		{
	    	putchar('-');
			x=-x;
		}
	    if(x>9) write(x/10);
	    putchar(x%10+'0');
	}
}
using namespace my_std;
const ll N=1e5+50;
ll n,w,a[N],b[N],s1[N],s2[N],ans1,ans2;
inline bool ck1(int pos,int sum){return s2[n]-s2[n-pos]>=sum;}
inline bool ck2(int pos,int sum){return s1[n]-s1[n-pos]>=sum;} 
int main(void)
{
	n=read(),w=read();
	fr(i,1,n) a[i]=read();
	fr(i,1,n) b[i]=read();
//	reverse(a+1,a+n+1);
//	reverse(b+1,b+n+1);
	fr(i,1,n) s1[i]=s1[i-1]+a[i];
	fr(i,1,n) s2[i]=s2[i-1]+b[i];
    fr(i,1,n)
	{
		ll sum=s1[n]-s1[n-i],l=1,r=n;
		while(l<=r)
		{
			ll mid=(l+r)>>1;
			if(ck1(mid,sum))
			{
				ans1=max(ans1,sum-(i+mid)*w);
				r=mid-1;
			}
			else l=mid+1;
		}
	}
	fr(i,1,n)
	{
		ll sum=s2[n]-s2[n-i],l=1,r=n;
		while(l<=r)
		{
			ll mid=(l+r)>>1;
			if(ck2(mid,sum))
			{
				ans2=max(ans2,sum-(i+mid)*w);
				r=mid-1;
			}
			else l=mid+1;
		}
	}
	write(max(ans1,ans2));
	putchar('
');
    return 0;
}