HDOJ 5461 Largest Point（沈阳网络赛）

Largest Point

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 548    Accepted Submission(s): 234


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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.
 

[align=left]Output[/align]
The output contains exactly
T
lines.

For each test case, you should output the maximum value of
at2i+btj.
 

[align=left]Sample Input[/align]

2

3 2 1
1 2 3

5 -1 0
-3 -3 0 3 3

 

[align=left]Sample Output[/align]

Case #1: 20
Case #2: 0

分情况讨论，代码当时写的，下去了也没改，有点丑

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#define INF 0x7fffffffff;
using namespace std;
int num[5000100];
int v[5000100];
int main()
{
int t,k;
long long sum;
int i,j,a,b,n;
long long mi,M;
int cas=0;
scanf("%d",&t);
while(t--)
{
sum=0;
scanf("%d%d%d",&n,&a,&b);
for(i=0;i<n;i++)
scanf("%d",&num[i]);
memset(v,0,sizeof(v));
if(a<0)
{
mi=INF;
for(i=0;i<n;i++)
{
if(fabs(num[i])<mi)
{
k=i;
mi=fabs(num[i]);
}
}
sum+=(a*mi*mi);
v[k]=1;
}
else
{
if(a>0)
{
M=-INF;
for(i=0;i<n;i++)
{
if(fabs(num[i])>M)
{
M=fabs(num[i]);
k=i;
}
}
sum+=(a*M*M);
v[k]=1;
}
}
if(b<0)
{
mi=INF;
for(i=0;i<n;i++)
{
if(num[i]<mi&&!v[i])
{
mi=num[i];
}
}
sum+=(b*mi);
}
else
{
if(b>0)
{
M=-INF;
for(i=0;i<n;i++)
{
if(num[i]>M&&!v[i])
{
M=num[i];
}
}
sum+=(b*M);
}
}
printf("Case #%d: %lld\n",++cas,sum);
}
return 0;
}