Largest Point

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5461

Description

Given the sequence A with n integers t1,t2,⋯,tn. Given the integral coefficients a and b. The fact that select two elements ti and tj of A and i≠j to maximize the value of at2i+btj, becomes the largest point.

Input

An positive integer T, indicating there are T test cases.
For each test case, the first line contains three integers corresponding to n (2≤n≤5×106), a (0≤|a|≤106) and b (0≤|b|≤106). The second line contains n integers t1,t2,⋯,tn where 0≤|ti|≤106 for 1≤i≤n.

The sum of n for all cases would not be larger than 5×106.

Output

The output contains exactly T lines.
For each test case, you should output the maximum value of at2i+btj.

Sample Input

2

3 2 1
1 2 3

5 -1 0
-3 -3 0 3 3

Sample Output

Case #1: 20
Case #2: 0

HINT

题意

给你a,b，再给你n个数

然后让你求ati*ti+btj最大值是多少

题解：

我们是暴力做的，找到最大的两个数，最小的两个数，绝对值最大的两个数，绝对值最小的两个数

然后扔进一个vector里面，然后去重，然后暴力枚举的

但是还是怕tle，就分治了一下解法，数据小的话就直接n^2暴力枚举= =

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <queue>
#include <iomanip>
#include <string>
#include <ctime>
#include <list>
#include <bitset>
typedef unsigned char byte;
#define pb push_back
#define input_fast std::ios::sync_with_stdio(false);std::cin.tie(0)
#define local freopen("in.txt","r",stdin)
#define pi acos(-1)

using namespace std;
const int maxn = 5e6 + 500;
long long  a , b  , ans , p[maxn];
int vis[11] , used[maxn] , n , sz;
vector<long long>s;

struct data
{
    long long val;
    int idx;
};

data A[maxn] , B[maxn];



bool cmp1(const data & x,const data & y)
{
    return x.val < y.val;
}

bool cmp2(const data & x,const data & y)
{
    return abs(x.val) < abs(y.val);
}


void initiation()
{
    scanf("%d%I64d%I64d",&n,&a,&b);
    memset(used,0,sizeof(int)*(n+2));
    for(int i = 0 ; i < n ; ++ i)
    {
        scanf("%I64d",&A[i].val);
        A[i].idx = i;
        B[i].val = A[i].val;
        B[i].idx = i;
        p[i] = A[i].val;
    }
    s.clear();
    ans = -(1LL<<58);
}

void dfs(int cur , long long check)
{
    if(cur == 2) ans = max( ans , check);
    else
    {
        for(int i = 0 ; i < sz ; ++ i) 
        {
            if(!vis[i])
            {
                vis[i] = 1;
                if(cur == 0) dfs(cur + 1 , check + a * s[i]*s[i] );
                else dfs(cur + 1 , check + b*s[i]);
                vis[i] = 0;
            }
        }
    }
}



long long solve()
{
    sort(A,A+n,cmp1);
    sort(B,B+n,cmp2);
    used[A[0].idx] = 1 , used[A[1].idx] = 1 , used[A[n-1].idx] = 1 , used[A[n-2].idx] = 1;
    used[B[0].idx] = 1 , used[B[1].idx] = 1 ; used[B[n-1].idx] = 1 , used[B[n-2].idx] = 1;
    for(int i = 0 ; i < n ; ++ i) if(used[i]) s.push_back(p[i]);
    sz = s.size();
    memset(vis,0,sizeof(vis));
    dfs(0,0);
    return ans;
}

long long solve2()
{
    for(int i=0;i<n;i++)
     {
        for(int j=0;j<n;j++)
            {
                if(i==j) continue;
                ans = max(ans,A[i].val*A[i].val*a+b*A[j].val);
            }
        }
     return ans;
}

int main(int argc,char *argv[])
{
    int Case;
    scanf("%d",&Case);
    for(int cas = 1 ; cas <= Case ; ++ cas)
    {
        initiation();
        printf("Case #%d: ",cas);
        if(n<=70) printf("%I64d\n",solve2());
        else printf("%I64d\n",solve());
    }
    return 0;
}