hdu 5461 Largest Point(暴搞)-优快云博客

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Largest Point

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1450 Accepted Submission(s): 579

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
  
  

solution:
	求a*ti^2+b*tj 且i!=j 直接暴力即可 记录前者和后者最大和次大 用map统计一下最大出现的次数 然后瞎搞即可
#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;
typedef long long ll;
int n;
ll a, b,x,y,z;
int main()
{
    int t,cnt=1;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%I64d%I64d", &n, &a, &b);
        ll max1 = -1e18, max2 = -1e18 - 1, max3 = -1e18, max4 = -1e18 - 1;
        int pos1, pos3;
        map<long long, int>q;
        for (int i = 0; i < n; i++)
        {
            scanf("%I64d", &x);
            y = a*x*x; z = b*x;
            if (y > max1)
            {
                max2 = max1;
                max1 = y; pos1 = i;
                q[max1]++;
            }
            else if (y > max2)
            {
                max2 = y;
            }
            if (z > max3)
            {
                max4 = max3;
                max3 = z; pos3 = i;
                q[max3]++;
            }
            else if (z > max4)
            {
                max4 = z;
            }
        }
        printf("Case #%d: ", cnt++);
        if (pos1 != pos3 || q[max1] > 1||q[max3]>1)printf("%I64d\n", max1 + max3);
        else printf("%I64d\n", max(max1 + max4, max2 + max3));
    }
    return 0;
}