Largest Point
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 474 Accepted Submission(s): 213
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 nintegers 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.
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 nintegers 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.
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
题目大意：有T组数据，然后第一行有三个数n, a, b;
接下来有n行arr[i],求a * arr[i] + b * arr[j] 的最大值（i != j）
解题思路:
分四种情况讨论，有点麻烦，但是却管用，先对arr数组排一下序，
然后在对绝对值数组data排一下序，在判断一下是不是绝对值数组的最值跟原先数组的最值一样，
最后在确定是不是有多个最值就okl
上代码：
/**
2015 - 09 - 20 中午
Author: ITAK
Motto:
今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef long long LL;
const int maxn = 5*1e6+5;
const double eps = 1e-7;
LL arr[maxn], data[maxn];
int main()
{
int T;
scanf("%d",&T);
for(int cas=1; cas<=T; cas++)
{
int m, a, b;
scanf("%d%d%d",&m,&a,&b);
for(int i=0; i<m; i++)
{
scanf("%lld",&arr[i]);
data[i] = abs(arr[i]);
}
sort(data, data+m);
sort(arr, arr+m);
LL ans;
if(a < 0)
{
if(b <= 0)
{
if(abs(arr[0]) != data[0])
{
ans = data[0]*data[0]*a + arr[0]*b;
}
else
{
if(data[0] != data[1])
ans = data[0]*data[0]*a + arr[1]*b;
else
ans = data[0]*data[0]*a + arr[0]*b;
}
}
else
{
if(abs(arr[m-1]) != data[0])
{
ans = data[0]*data[0]*a + arr[m-1]*b;
}
else
{
if(data[0] != data[1])
ans = data[0]*data[0]*a + arr[m-2]*b;
else
ans = data[0]*data[0]*a + arr[m-1]*b;
}
}
}
else
{
if(b <= 0)
{
if(abs(arr[0]) != data[m-1])
{
ans = data[m-1]*data[m-1]*a + arr[0]*b;
}
else
{
if(data[m-1] != data[m-2])
ans = data[m-1]*data[m-1]*a + arr[1]*b;
else
ans = data[m-1]*data[m-1]*a + arr[0]*b;
}
}
else
{
if(abs(arr[m-1]) != data[m-1])
{
ans = data[m-1]*data[m-1]*a + arr[m-1]*b;
}
else
{
if(data[m-1] != data[m-2])
ans = data[m-1]*data[m-1]*a + arr[m-2]*b;
else
ans = data[m-1]*data[m-1]*a + arr[m-1]*b;
}
}
}
printf("Case #%d: %lld\n",cas, ans);
}
return 0;
}
