HDU----5461水题

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Description

Given the sequence

with

integers

. Given the integral coefficients

and

. The fact that select two elements

and

to maximize the value of

, becomes the largest point.

Input

An positive integer

, indicating there are

test cases.
For each test case, the first line contains three integers corresponding to

and

. The second line contains

integers

where

for

.

The sum of

for all cases would not be larger than

Output

The output contains exactly

lines.
For each test case, you should output the maximum value of

Sample Input

23 2 11 2 35 -1 0-3 -3 0 3 3

Sample Output

Case #1: 20Case #2: 0
#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<algorithm>#include<iostream>using namespace std;#define N 5000100#define INF 0x3f3f3f3flong long s[N];struct node{    long long sum, k;}w[N], p[N];int cmp(node a, node b){    return a.sum<b.sum;}int main(){    int i, n, a, b, T, t=1;    scanf("%d", &T);    while(T--)    {        scanf("%d%d%d", &n,&a,&b);        for(i=0;i<n;i++)        {            scanf("%lld", &s[i]);            w[i].sum=a*s[i]*s[i];            w[i].k=s[i];            p[i].sum=b*s[i];            p[i].k=s[i];        }        sort(w, w+n, cmp);        sort(p, p+n, cmp);        printf("Case #%d: ",t++);        if(w[n-1].k!=p[n-1].k)            printf("%lld\n", w[n-1].sum+p[n-1].sum);        else        {            long long ans=max(w[n-1].sum+p[n-2].sum,w[n-2].sum+p[n-1].sum);            printf("%lld\n",ans);        }    }    return 0;}

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