考虑两种情况

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Largest Point

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 795 Accepted Submission(s): 327

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
模拟太难想了，种类繁多。要快速解决这问题，直接暴力，只要考虑a,b,的值就行了。。。
#include<iostream>#include<algorithm>#include<cmath> using namespace std;long long a[5000001];long long ans1,ans2,s1,s2,ans3,ans4;int main(){int n,t,ca=1,o;long long m,k;    scanf("%d",&t);while(t--){ans1=ans2=ans3=ans4=0;scanf("%d%I64d%I64d",&n,&m,&k);for(int i=1;i<=n;i++){scanf("%I64d",&a[i]);}//sort(a+1,a+1+n);s1=LLONG_MIN;for(int i=1;i<=n;i++){s2=m*a[i]*a[i];if(s1<s2){s1=s2;o=i; } }ans1=s1;s1=LLONG_MIN; for(int i=1;i<=n;i++){if(i!=o){s2=a[i]*k;s1=max(s1,s2);}}ans2=s1;    s1=LLONG_MIN;for(int i=1;i<=n;i++){s2=a[i]*k;if(s1<s2){s1=s2;o=i; } }ans3=s1;s1=LLONG_MIN; for(int i=1;i<=n;i++){if(i!=o){s2=m*a[i]*a[i];s1=max(s1,s2);}}ans4=s1;ans1=ans1+ans2;ans3=ans3+ans4;ans4=max(ans1,ans3);printf("Case #%d: %I64d\n",ca++,ans4);}return 0; }

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