hdu5461 Largest Point（沈阳网赛）

Largest Point

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 536 Accepted Submission(s): 230

Problem 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

23 2 11 2 35 -1 0-3 -3 0 3 3

Sample Output

Case #1: 20Case #2: 0

Source

2015 ACM/ICPC Asia Regional Shenyang Online 

题意：求a*t1*t1+b*t2的值最大。

分析：数据不大，可以直接暴力求解，详解见代码。

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <cmath>#include <algorithm>using namespace std;const double eps = 1e-6;const double pi = acos(-1.0);const int INF = 0x3f3f3f3f;const int MOD = 1000000007;#define ll long long#define CL(a) memset(a,0,sizeof(a))ll T,n,a,b;ll t[1000010];ll min1,min2,max1,max2,k;//分别存最小的数、第二小的数、最大的数、第二大的数和最接近0的数int main (){    scanf ("%lld",&T);    for (int cas=1; cas<=T; cas++)    {        scanf ("%lld%lld%lld",&n,&a,&b);        k=INF;        for (int i=0; i<n; i++)        {            scanf ("%lld",&t[i]);        }        cout<<"Case #"<<cas<<": ";        sort(t, t+n);        for (int i=0; i<n; i++)        {            if (t[i]<=0&&t[i+1]>=0)                k=min(-t[i], t[i+1]);        }        min1=t[0]; min2=t[1];        max1=t[n-1]; max2=t[n-2];//找出这五个数        if (a<0&&b<0)//然后就是苦逼的找最大解了，注意负数的平方为正数        {            printf ("%lld\n",a*k*k+b*min1);        }        else if (a<0&&b>0)        {            printf ("%lld\n",a*k*k+b*max1);        }        else if (a>0&&b<0)        {            printf ("%lld\n",max(max(a*max1*max1+b*min1, a*min1*min1+b*min2), a*min2*min2+b*min1));        }        else if (a>0&&b>0)        {            printf ("%lld\n",max(max(a*max1*max1+b*max2, a*max2*max2+b*max1), a*min1*min1+b*max1));        }        else printf ("0\n");    }    return 0;}