2019牛客暑期多校训练营（第四场）C.sequence

2019牛客暑期多校训练营（第四场）C.sequence

题目链接

题目描述

Your are given two sequences $a_{1\dots n}$ and $b_{1\dots n}$.You need to answer $\underset{1\le l\le r\le n}{\max }\{min(a_{l\dots r})\times sum(b_{l\dots r})\}$ 。

Where min(a) means the minimal value of every element of sequence a, sum(a) means the sum of every element of sequence a .

输入描述:

The first line contains an integer n .

The second line contains n integers meaning $a_{1\dots n}$.

The third line contains n integers meaning $b_{1\dots n}$.

输出描述:

An integer meaning the answer.

示例1

输入

3
1 -1 1
1 2 3

输出

3

这题可以直接遍历求答案，对 $a[i]>0$ 时，要使 $sum$ 尽可能大，如果 $a[i]<0$，就要使 $sum$ 尽可能小，对于小于 0 的情况，我们可以将所有数取反再算一次，避免遗漏最大值，AC代码如下：

#include
using namespace std;
typedef long long ll;
const int N=3e6+5;
int n;
ll a[N],b[N];
int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++) scanf("%lld",&a[i]);
    for(int i=0;i<n;i++) scanf("%lld",&b[i]);
    ll sum=b[0],ans=a[0]*b[0],now=a[0];
    for(int i=1;i<n;i++){
        ans=max(ans,b[i]*a[i]);
        if(sum<0){
            sum=b[i],now=a[i];
        }else{
            sum+=b[i];
            now=min(now,a[i]);
            ans=max(ans,now*sum);
        }
        a[i]*=-1,b[i]*=-1;
    }
    sum=b[0],now=a[0];
    for(int i=1;i<n;i++){
        if(sum<0){
            sum=b[i],now=a[i];
        }else{
            sum+=b[i];
            now=max(now,a[i]);
            ans=max(ans,now*sum);
        }
    }
    printf("%lld",ans);
}