upc 8262: LARGEST POINT

题目描述

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 becomes the largest point.

输入

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 .

输出

The output contains exactly T lines. For each test case, you should output the maximum value of

样例输入

2
3 2 1
1 2 3
5 -1 0
-3 -3 0 3 3

样例输出

Case #1: 20
Case #2: 0

题意：计算a*ti*ti+b*tj的最大值，且j != i

思路：分别计算a*t*t的值和b*t的值，并记录t的下标，比较两者最大值下标是否相等，不等直接输出结果，相等再比较次

最大值，输出结果。。。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef struct node
{ll step;//记录下标ll value;//记录值
}p;
p maxx[1000010],mbxx[1000010];//分别存a*t*t和b*t的值和下标
bool cmp(p a,p b)
{return a.value>b.value;
}
int main()
{int t;int temp=1;scanf("%d",&t);while(t--){ll n,a,b;scanf("%lld%lld%lld",&n,&a,&b);for(int i=0;i<n;i++){ll x;scanf("%lld",&x);maxx[i].step=i;maxx[i].value=a*x*x;mbxx[i].step=i;mbxx[i].value=b*x;}sort(maxx,maxx+n,cmp);sort(mbxx,mbxx+n,cmp);if(maxx[0].step!=mbxx[0].step)//判断最大值下标printf("Case #%d: %lld\n",temp++,maxx[0].value+mbxx[0].value);elseprintf("Case #%d: %lld\n",temp++,maxx[0].value+mbxx[1].value>maxx[1].value+mbxx[0].value?maxx[0].value+mbxx[1].value:maxx[1].value+mbxx[0].value);}return 0;
}

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