hdu 5461(分类讨论)

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 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

 

Source

2015 ACM/ICPC Asia Regional Shenyang Online

 

昨天输的很惨，中南地区邀请赛只A了两道题。所以感觉要多做少错思路题。

这个题分4种情况考虑，每种又分为三种情况。举一个例子：

a>0,b<0时

那么ti^2尽可能的大，tj尽可能的小。如果i！=j 那么直接取ti^2最大与tj最小。

如果i==j 那么在ti^2次大与tj最小和ti^2最大与tj次小中取大者。

还有就是 INF不能这样赋值 ，如1<<40 ,这个地方WA了好多次，，他会变成0

#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
",t++,ans);
    }
    return 0;
}