hdu5461(2015 ACM/ICPC Asia Regional Shenyang Online)

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

 

Source

2015 ACM/ICPC Asia Regional Shenyang Online

 

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 网络赛的签到题，排个序，然后在判断一下是否使用了同一个数，值得注意的是虽然a,b比较小，但需要定义成long long,因为t的平方会爆int；

代码

#include <iostream>#include <stdio.h>#include <algorithm>using namespace std;const int maxn=5000100;int t[maxn];struct hehe{  long long num;  int id;}aa[maxn],bb[maxn];long long cmp(hehe a,hehe b){ return a.num>b.num;}int main(){    int T,n,flag=0;    long long tmp1,tmp2,tmp,a,b;    scanf("%d",&T);    while(T--)    {     flag++;     scanf("%d%lld%lld",&n,&a,&b);     for(int i=0;i<=n-1;i++)      scanf("%d",&t[i]);     for(int i=0;i<=n-1;i++)     {      aa[i].num=a*t[i]*t[i];      aa[i].id=i;      bb[i].num=b*t[i];      bb[i].id=i;     }     sort(aa,aa+n,cmp);     sort(bb,bb+n,cmp);     if(aa[0].id!=bb[0].id)      printf("Case #%d: %lld\n",flag,aa[0].num+bb[0].num);     else     {     tmp1=aa[0].num+bb[1].num;     tmp2=aa[1].num+bb[0].num;     tmp=(tmp1>tmp2)?tmp1:tmp2;      printf("Case #%d: %lld\n",flag,tmp);     }    }    return 0;}