hdu5461 Largest Point（沈阳网赛）

Largest Point

Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 536 Accepted Submission(s): 230

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

题解：

ax2和bx分开考虑；

代码：

 #include<stdio.h>
 #include<string.h>
 #include<math.h>
 #define MAX(x,y)(x>y?x:y)
 #define F for(int i=0;i<n;i++)
 const int MAXN=;
 const int INF=0x3f3f3f3f;
 int m[MAXN],vis[MAXN];
 int main(){
     int T;
     int n,a,b,flot=;
     scanf("%d",&T);
     while(T--){
         memset(vis,,sizeof(vis));
         scanf("%d%d%d",&n,&a,&b);
         long long sum=,x,k;//xΪlong long
         F scanf("%d",m+i);
         if(a>){
             x=-INF;
             F if(x<fabs(m[i]))x=fabs(m[i]),k=i;
             vis[k]=;
             sum+=a*x*x;
         }
         else if(a<){
             x=INF;
                 F if(x>fabs(m[i]))x=fabs(m[i]),k=i;
                 vis[k]=;
             sum+=a*x*x;
         }
         if(b>){
              x=-INF;
             F if(x<m[i]&&!vis[i])x=m[i],k=i;
             vis[k]=;
             sum+=b*x;
         }
         else if(b<){
              x=INF;
                 F if(x>m[i]&&!vis[i])x=m[i],k=i;
                 vis[k]=;
             sum+=b*x;
         }
         printf("Case #%d: %lld\n",++flot,sum);
     }
     return ;
 }
 /*#include<stdio.h>
 #include<algorithm>
 #include<math.h>
 #define MAX(x,y)(x>y?x:y)
 #define js(x,y)(a*x*x+b*y)
 using namespace std;
 const int MAXN=5000010;
 const int INF=0x3f3f3f3f;
 int m[MAXN],ml[MAXN];
 int main(){
     int T,a,b,n,t[5],ans;
     scanf("%d",&T);
     for(int i=1;i<=T;i++){
         scanf("%d%d%d",&n,&a,&b);
         for(int j=0;j<n;j++)scanf("%d",m+j),ml[j]=fabs(m[j]);
             t[0]=*max_element(m,m+n);t[1]=*min_element(m,m+n);
             *max_element(m,m+n)=-INF;
             t[2]=*max_element(m,m+n);
             *min_element(m,m+n)=INF;
             *min_element(m,m+n)=INF;
             t[3]=*min_element(m,m+n);
             ans=-INF;
             printf("%d %d %d %d\n",t[0],t[1],t[2],t[3]);
             if(a>=0&&b>=0){
                 if(fabs(t[1])>=fabs(t[0]))
                     ans=js(t[1],t[0]);
                 else
                     ans=MAX(js(t[0],t[2]),js(t[2],t[0]));
             }
             else if(a>=0&&b<0){
                 if(fabs(t[0])>=fabs(t[1]))
                     ans=js(t[0],t[1]);
                 else
                     ans=MAX(js(t[1],t[3]),js(t[3],t[1]));
             }
             else{
                 int x=*min_element(ml,ml+n);

             }
             printf("Case #%d: %d\n",i,ans);
     }
     return 0;
 }*/