hdu5461-Largest Point(贪心)

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

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

Sample Output

Case #1: 20 Case #2: 0

给出t0-tn-1个数,和a,b,找出a*tx^2+b*ty的最大值,要求x!=y

这道题目数据有问题,根本没有500万个数,建一个50万的数组一样可以过

这里我们直接分别计算a*ti^2和b*ti,需要保持他们的序号,然后排序,去最大的两个,如果最大的两个序号相同,就

取 a*ti^2和b*ti分别第1大和第2大作比较,判断输出就可以了

AC:

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#define inf 9999999
#define maxn 5000007
using namespace std;int t,n,cas=1;
long long q;struct node
{long long x;int t;
}a[maxn],b[maxn];bool cmp(node x,node y)
{return x.x>y.x;
}int main()
{int cas=1,ai,bi;scanf("%d",&t);while(t--){scanf("%d%d%d",&n,&ai,&bi);for(int i=0;i<n;i++){scanf("%lld",&q);a[i].x=q*q*ai;a[i].t=i;b[i].x=q*bi;b[i].t=i;}printf("Case #%d: ",cas++);sort(a,a+n,cmp);sort(b,b+n,cmp);if(a[0].t!=b[0].t)printf("%lld\n",a[0].x+b[0].x);elseprintf("%lld\n",max((a[0].x+b[1].x),(a[1].x+b[0].x)));}return 0;
}

转载于:https://www.cnblogs.com/wangtao971115/p/10358312.html

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