Largest Point
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 548 Accepted Submission(s): 234
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
//要考虑s的最小绝对值与s的最大值相等或为同一个数时的情况。（在这块错了6次）
#include
#include
#include
#include
using namespace std;
#define INF 0x3f3f3f3f
int s[3000100];
int c[1000100];
int main()
{
int t;
int i,j,k,n,a,b;
int cnt=1;
long long int sum,mm;
scanf("%d",&t);
while(t--)
{
memset(c,0,sizeof(c));
sum=0;
scanf("%d%d%d",&n,&a,&b);
for(i=1;i<=n;i++)
scanf("%d",&s[i]);
printf("Case #%d: ",cnt++);
if(a>0)
{
mm=-INF;
for(i=1;i<=n;i++)
{
if(abs(s[i])>mm)
{
k=i;
mm=abs(s[i]);
}
}
c[k]=1;
sum+=a*mm*mm;
}
else
{
if(a<0)
{
mm=INF;
for(i=1;i<=n;i++)
{
if(abs(s[i])0)
{
mm=-INF;
for(i=1;i<=n;i++)
{
if(s[i]>mm&&!c[i])
{
mm=s[i];
}
}
sum+=b*mm;
}
else
{
if(b<0)
{
mm=INF;
for(i=1;i<=n;i++)
{
if(s[i]
#include
#include
#include
using namespace std;
int s[5000100];
int main()
{
int t,cnt=1;
int n,a,b,i,j;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&a,&b);
for(i=1;i<=n;i++)
scanf("%d",&s[i]);
printf("Case #%d: ",cnt++);
if(a>=0&&b>=0)
{
sort(s+1,s+n+1);
if(abs(s[1])>s[n])
printf("%lld\n",a*s[1]*s[1]+b*s[n]);
else
printf("%lld\n",a*s[n]*s[n]+b*s[n-1]>a*s[n-1]*s[n-1]+b*s[n]?a*s[n]*s[n]+b*s[n-1]:a*s[n-1]*s[n-1]+b*s[n]);
}
else if(a>0&&b<0)
{
sort(s+1,s+n+1);
if(abs(s[1])>s[n])
printf("%lld\n",a*s[1]*s[1]+b*s[2]>a*s[n]*s[n]+b*s[1]?a*s[1]*s[1]+b*s[2]:a*s[n]*s[n]+b*s[1]);
else
printf("%lld\n",a*s[n]*s[n]+b*s[1]);
}
else if(a<0&&b>0)
{
sort(s+1,s+n+1);
int mm=s[n];
int nn=s[n-1];
for(i=1;i<=n;i++)
s[i]=abs(s[i]);
sort(s+1,s+n+1);
if(mm!=s[1])
printf("%lld\n",a*s[1]*s[1]+b*mm);
else
printf("%lld\n",a*mm*mm+b*nn>a*s[2]*s[2]+b*mm?a*mm*mm+b*nn:a*s[2]*s[2]+b*mm);
}
else
{
sort(s+1,s+n+1);
int mm=s[1];
int nn=s[2];
for(i=1;i<=n;i++)
s[i]=abs(s[i]);
sort(s+1,s+n+1);
if(mm!=s[1])
printf("%lld\n",a*s[1]*s[1]+b*mm);
else
printf("%lld\n",a*mm*mm+b*nn>a*nn*nn+b*mm?a*mm*mm+b*nn:a*nn*nn+b*mm);
}
}
return 0;
}
