Largest Point

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)

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

2

3 2 1
1 2 3

5 -1 0
-3 -3 0 3 3

Sample Output

Case #1: 20
Case #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;
}

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