2024.12.25 周三

2024.12.25 周三

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Q1. 1100

A subarray is a continuous part of array.

Yarik recently found an array \(a\) of \(n\) elements and became very interested in finding the maximum sum of a non empty subarray. However, Yarik doesn't like consecutive integers with the same parity, so the subarray he chooses must have alternating parities for adjacent elements.

For example, \([1, 2, 3]\) is acceptable, but \([1, 2, 4]\) is not, as \(2\) and \(4\) are both even and adjacent.

You need to help Yarik by finding the maximum sum of such a subarray.

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------------------------独自思考分割线------------------------

• 思路好想，实现的时候出了很多问题。

A1.

1. 首先如果不考虑额外条件。一眼：遍历求前缀和同时维护最小前缀和(不能为空所以是\(i-1\)的)，同时更新答案 \(pre[i]-min\_pre\) 。
2. 额外条件是连续子数组不包含连续奇偶性相同的数。简单思考发现以满足额外条件将数组分为多个连续子数组，每个子数组单独计算。即分割后互不干扰/贡献独立。

------------------------代码分割线------------------------

A1.

#include <bits/stdc++.h>
#define int long long
using namespace std;

void _();
signed main()
{
   int t=1;cin>>t;
   while(t--) _();
}

void _()
{
    int n;cin>>n;
    vector<int> a(n+1);
    for(int i=1;i<=n;i++) cin>>a[i];
    vector<vector<int>> vec;
    for(int i=1;i<=n;i++)
    {
        vector<int> t{a[i]};
        int j=i+1;
        for(;j<=n&&((abs(a[j])%2)!=(abs(a[j-1])%2));j++) // &1优先级、对负数   负数%2
        t.push_back(a[j]);
        vec.push_back(t);
        i=j-1;
    }
    int res=-1e12;
    for(auto a:vec)
    {
        // for(auto v:a) cout<<v<<' ';cout<<endl;
        int ans=0; 
        res=max(res,a[0]);
        if(a.size()>1)
        {
            ans=min(ans,a[0]);
            int pre=a[0];
            for(int i=1;i<a.size();i++)
            {  ans=min(ans,pre);
                pre+=a[i]; 
                res=max(res,pre-ans);
              
            }
        }
    }
    cout<<res<<endl;
}

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