hdu 5461 Largest Point（杂题）

最新推荐文章于 2019-05-03 20:33:03 发布

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最新推荐文章于 2019-05-03 20:33:03 发布

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分类专栏：杂题文章标签： ACM HDU

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本文介绍了一种算法问题——最大点值问题，该问题要求在给定整数序列中找到两个不同元素，使得特定组合的数学表达式值最大化。文章详细阐述了解决方案的思路，并提供了实现代码。

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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

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

题意，给你n个点和a，b，找出两个不同的点使得at2i+btj.的值最大

思路：分四种情况讨论即可。

代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
using namespace std;
typedef long long LL;
#define INF 999999999
int main(){
    int tcase;
    scanf("%d",&tcase);
    int t =1;
    while(tcase--){
        LL IMAX=-INF,ISMAX=-INF,IMIN=INF,ISMIN=INF; ///ti平方的最大次大最小次小
        LL JMAX=-INF,JSMAX=-INF,JMIN=INF,JSMIN=INF; ///tj的最大次大最小次小
        int n;
        LL a,b;
        int id1,id2,id3,id4; ///ti 最大最小 tj最大最小
        scanf("%d%lld%lld",&n,&a,&b);
        for(int i=1;i<=n;i++){
            LL NUM,TEMP;
            scanf("%lld",&NUM);
            TEMP = NUM*NUM;
            if(NUM<JMIN){
                JSMIN = JMIN;
                JMIN = NUM;
                id2 = i;
            }else JSMIN = min(JSMIN,NUM);
            if(NUM>JMAX){
                JSMAX = JMAX;
                JMAX = NUM;
                id1 = i;
            }else JSMAX = max(JSMAX,NUM);
            if(TEMP<IMIN){
                ISMIN = IMIN;
                IMIN = TEMP;
                id4 = i;
            }else ISMIN = min(ISMIN,TEMP);
            if(TEMP>IMAX){
                ISMAX = IMAX;
                IMAX = TEMP;
                id3 = i;
            }else ISMAX = max(ISMAX,TEMP);
        }
        LL ans = 0;
        if(a>=0&&b>=0){
            if(id3!=id1) ans = a*IMAX+b*JMAX;
            else ans = max(a*ISMAX+b*JMAX,a*IMAX+b*JSMAX);
        }else if(a>0&&b<0){
            if(id3!=id2) ans = a*IMAX+b*JMIN;
            else ans = max(a*ISMAX+b*JMIN,a*IMAX+b*JSMIN);
        }else if(a<0&&b>0){
            if(id4!=id1) ans = a*IMIN+b*JMAX;
            else ans = max(a*ISMIN+b*JMAX,a*IMIN+b*JSMAX);
        }else {
            if(id4!=id2) ans = a*IMIN+b*JMIN;
            else ans = max(a*IMIN+b*JSMIN,a*ISMIN+b*JMIN);
        }
        printf("Case #%d: %lld\n",t++,ans);
    }
    return 0;
}