\(Solution:\)
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('\n');
return 0;
}