Largest Point

Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 1485    Accepted Submission(s): 588

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 ij 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 (2n5×106), a (0|a|106) and b (0|b|106). The second line contains n integers t1,t2,,tn where 0|ti|106 for 1in.

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

a>0,b<0时

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL INF = 1<<60;
int main(){
int tcase;
scanf("%d",&tcase);
int t =1;
while(tcase--){
int n,a,b;
scanf("%d%d%d",&n,&a,&b);
LL ans;
if(a>=0&&b>=0){
LL fmax = -INF,smax = -INF;
for(int i=1;i<=n;i++){
LL num;
scanf("%lld",&num);
if(num>fmax){
smax = fmax;
fmax = num;
}else{
smax = max(smax,num);
}
}
ans = max(a*fmax*fmax+b*smax,a*smax*smax+b*fmax);
}else if(a>0&&b<0){
LL MAX=-INF,SMAX=-INF; ///平方的最大和次大
LL MIN=INF,SMIN=INF; ///最小和次小
int id1,id2;
for(int i=1;i<=n;i++){
LL num;
scanf("%lld",&num);
LL temp = num*num;
if(num<MIN){
SMIN = MIN;
MIN = num;
id1 = i;
}else SMIN = min(SMIN,num);
if(temp>MAX){
SMAX = MAX;
MAX  = temp;
id2 = i;
}else SMAX = max(SMAX,num);
}
if(id1!=id2){
ans = a*MAX+b*MIN;
}else{
ans = max(a*SMAX+b*MIN,a*MAX+b*SMIN);
}
}else if(a<0&&b>0){
LL MAX = -INF,MIN = INF; ///注意这里的最小是指平方的最小
LL smin=INF,smax = -INF; ///此处次小是指平方的次小
int id1,id2; ///分别记录ti 和 tj 的位置
for(int i=1;i<=n;i++){
LL num;
scanf("%lld",&num);
if(num>MAX) {
smax = MAX;
MAX = num;
id1 = i;
}else smax = max(smax,num);
LL temp = num*num;
if(temp<MIN){
smin = MIN;
MIN = temp;
id2 = i;
}else smin = min(smin,temp);
}
if(id1!=id2){
ans = a*MIN+b*MAX;
}
else{
ans = max(a*MIN+b*smax,a*smin+b*MAX);
}
}else {
LL MIN = INF,SMIN = INF; ///这两个值代表绝对值次小和最小的平方
LL MIN1 = INF,SMIN1 = INF;  ///这两个值代表次小和最小
int id1,id2;
for(int i=1;i<=n;i++){
LL num;
scanf("%lld",&num);
LL temp = num*num;
if(num<MIN1) {
SMIN1 = MIN1;
MIN1 = num;
id1 = i;
}else SMIN1 = min(SMIN1,num);

if(temp<MIN){
SMIN = MIN;
MIN = temp;
id2 = i;
}else SMIN = min(SMIN,temp);
}
if(id1!=id2){
ans = a*MIN+b*MIN1;
}else{
ans = max(a*MIN+b*SMIN1,a*SMIN+b*MIN1);
}
}
printf("Case #%d: %lld\n",t++,ans);
}
return 0;
}