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.
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.
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
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
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<math.h> 5 #include<algorithm> 6 7 using namespace std; 8 9 const int maxn = 5*1e6+7; 10 #define INF 0x3f3f3f3f 11 12 struct node 13 { 14 int x, id; 15 }P[maxn]; 16 17 int cmp(node a, node b) 18 { 19 return abs(a.x) < abs(b.x); 20 } 21 22 int cmp1(node a, node b) 23 { 24 return a.x < b.x; 25 } 26 27 int main() 28 { 29 int t, n, a, b, l = 1; 30 int A1, a1, ma, B1, b1, mb; 31 scanf("%d", &t); 32 long long sum, ans; 33 while(t--) 34 { 35 scanf("%d%d%d", &n, &a, &b); 36 for(int i = 0; i < n; i++) 37 { 38 scanf("%d", &P[i].x); 39 P[i].id = i; 40 } 41 A1 = a1 = ma = B1 = b1 = mb = 0; 42 sort(P, P+n, cmp); 43 if(a >= 0) 44 A1 = P[n-1].x, a1 = P[n-1].id, ma = P[n-2].x; 45 else 46 A1 = P[0].x, a1 = P[0].id, ma = P[1].x; 47 sort(P, P+n, cmp1); 48 if(b >= 0) 49 B1 = P[n-1].x, b1 = P[n-1].id, mb = P[n-2].x; 50 else 51 B1 = P[0].x, b1 = P[0].id, mb = P[1].x; 52 if(a1 != b1) 53 { 54 sum = (long long)a * A1 *A1; 55 sum += (long long)b * B1; 56 } 57 58 else 59 { 60 sum = (long long)a * A1*A1; 61 sum += (long long)b*mb; // 为什么非得这么加，这么强制转换，orWA 62 ans = (long long)a*ma*ma; 63 ans += (long long)b*B1; 64 sum = sum > ans ? sum : ans; 65 } 66 printf("Case #%d: %I64d ",l++, sum); 67 } 68 return 0; 69 }