Largest Point

Largest Point

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 969    Accepted Submission(s): 384

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 nintegers 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

Source

2015 ACM/ICPC Asia Regional Shenyang Online

题意：在一个数组里找两个下标不一样的数，使at2i+btj最大，下标不一样，数值可能一样

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<math.h>
 #include<algorithm>

 using namespace std;

 const int maxn = *1e6+;
 #define INF 0x3f3f3f3f

 struct node
 {
     int x, id;
 }P[maxn];

 int cmp(node a, node b)
 {
     return abs(a.x) < abs(b.x);
 }

 int cmp1(node a, node b)
 {
     return a.x < b.x;
 }

 int main()
 {
     int t, n, a, b, l = ;
     int A1, a1, ma, B1, b1, mb;
     scanf("%d", &t);
     long long sum, ans;
     while(t--)
     {
         scanf("%d%d%d", &n, &a, &b);
         for(int i = ; i < n; i++)
         {
             scanf("%d", &P[i].x);
             P[i].id = i;
         }
         A1 = a1 = ma = B1 = b1 = mb = ;
         sort(P, P+n, cmp);
         if(a >= )
             A1 = P[n-].x, a1 = P[n-].id, ma = P[n-].x;
         else
             A1 = P[].x, a1 = P[].id, ma = P[].x;
         sort(P, P+n, cmp1);
         if(b >= )
             B1 = P[n-].x, b1 = P[n-].id, mb = P[n-].x;
         else
             B1 = P[].x, b1 = P[].id, mb = P[].x;
         if(a1 != b1)
         {
             sum = (long long)a * A1 *A1;
             sum += (long long)b * B1;
         }

         else
         {
             sum = (long long)a * A1*A1;
             sum += (long long)b*mb;   // 为什么非得这么加，这么强制转换，orWA
             ans = (long long)a*ma*ma;
             ans += (long long)b*B1;
             sum = sum > ans ? sum : ans;
         }
         printf("Case #%d: %I64d\n",l++,  sum);
     }
     return ;
 }