zoukankan      html  css  js  c++  java
  • HDU

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5461

    题意:求a*c[i]*c[i]+b*c[j]的最大值;

    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

    先贴一个还不明白该怎么改的代码:

      1 #include<iostream>
      2 #include<algorithm>
      3 #include<cstdio>
      4 #include<cstring>
      5 #include<queue>
      6 #include<stdlib.h>
      7 #include<map>
      8 #include<cmath>
      9 
     10 using namespace std;
     11 
     12 #define N 500010
     13 #define INF 0x3f3f3f3f
     14 
     15 long long sum[N];
     16 
     17 struct node
     18 {
     19     long long s,o;
     20 }p[N];
     21 
     22 bool cmp1(node a,node b)
     23 {
     24     return a.s<b.s;
     25 }
     26 
     27 bool cmp2(node a,node b)
     28 {
     29     return a.o<b.o;
     30 }
     31 
     32 int main()
     33 {
     34     int T,k=1,i,j;
     35     long long n,a,b,x,y,minn,minn2,maxx,maxx2,minabs,maxabs;
     36 
     37     scanf("%d", &T);
     38 
     39     while(T--)
     40     {
     41         scanf("%lld %lld %lld", &n,&a,&b);
     42         memset(sum,-INF,sizeof(sum));
     43 
     44         for(i=0;i<n;i++)
     45         {
     46             scanf("%lld", &p[i].s);
     47             p[i].o=abs(p[i].s);
     48         }
     49 
     50         sort(p,p+n,cmp1);///数从小到大排序
     51         minn=p[0].s;
     52         minn2=p[1].s;
     53         maxx=p[n-1].s;
     54         maxx2=p[n-2].s;
     55         sort(p,p+n,cmp2);///绝对值排序
     56         minabs=p[0].s;
     57         maxabs=p[n-1].s;
     58 
     59         i=0;
     60         if(a>=0)
     61         {
     62             x=maxabs;
     63             if(b>=0)
     64             {
     65                 if(maxx==x)
     66                 {
     67                     y=maxx2;
     68                     sum[i++]=max(a*x*x+b*y,a*y*y+b*x);
     69                 }
     70                 else
     71                 {
     72                     y=maxx;
     73                     sum[i++]=a*x*x+b*y;
     74                 }
     75 
     76             }
     77             else
     78             {
     79                 if(minn==x)
     80                 {
     81                     y=minn2;
     82                     sum[i++]=max(a*x*x+b*y,a*y*y+b*x);
     83                 }
     84                 else
     85                 {
     86                      y=minn;
     87                      sum[i++]=a*x*x+b*y;
     88                 }
     89 
     90             }
     91         }
     92         else
     93         {
     94             x=minabs;
     95             if(b>=0)
     96             {
     97                 if(maxx==x)
     98                 {
     99                     y=maxx2;
    100                     sum[i++]=max(a*x*x+b*y,a*y*y+b*x);
    101                 }
    102                 else
    103                 {
    104                     y=maxx;
    105                     sum[i++]=a*x*x+b*y;
    106                 }
    107 
    108             }
    109            else
    110             {
    111                 if(minn==x)
    112                 {
    113                     y=minn2;
    114                     sum[i++]=max(a*x*x+b*y,a*y*y+b*x);
    115                 }
    116                 else
    117                 {
    118                     y=minn;
    119                     sum[i++]=a*x*x+b*y;
    120                 }
    121 
    122             }
    123         }
    124 
    125         long long maxsum=-INF;
    126         for(j=0;j<i;j++)
    127             maxsum=max(maxsum,sum[j]);
    128 
    129         printf("Case #%d: %lld
    ",k++,maxsum);
    130     }
    131     return 0;
    132 }
    133 /*HDU 5461*/

    AC代码:直接暴力

     1 #include<iostream>
     2 #include<algorithm>
     3 #include<cstdio>
     4 #include<cstring>
     5 #include<queue>
     6 #include<stdlib.h>
     7 #include<map>
     8 #include<cmath>
     9 
    10 using namespace std;
    11 
    12 #define N 5100000
    13 #define INF 0x3f3f3f3f
    14 
    15 long long c[N];
    16 
    17 int main()
    18 {
    19     int T,k=1,i;
    20     long long n,a,b,x,y,minn,minn2,maxx,maxx2,maxsum,sum;
    21 
    22     scanf("%d", &T);
    23 
    24     while(T--)
    25     {
    26         scanf("%lld %lld %lld", &n,&a,&b);
    27         x=INF;
    28 
    29         for(i=0;i<n;i++)
    30             scanf("%lld", &c[i]);
    31 
    32         sort(c,c+n);
    33 
    34         for(i=0;i<n;i++)
    35         {
    36             if(c[i]<=0&&c[i+1]>=0)
    37                 x=min(-c[i],c[i+1]);
    38             ///对于全体绝对值排序找最小的情况来说,这样会更方便,反正其他的值也用不上
    39         }
    40 
    41         minn=c[0];
    42         minn2=c[1];
    43         maxx=c[n-1];
    44         maxx2=c[n-2];
    45 
    46 
    47         if(a<0&&b<0)
    48         {
    49             y=minn;
    50             maxsum=a*x*x+b*y;
    51         }
    52         else if(a<0&&b>0)
    53         {
    54              y=maxx;
    55              maxsum=a*x*x+b*y;
    56         }
    57         else if(a>0&&b>0)
    58         {
    59             x=maxx;
    60             y=maxx2;
    61             sum=max(a*x*x+b*y,a*y*y+b*x);
    62             maxsum=max(sum,a*minn*minn+b*maxx);
    63         }
    64         else if(a>0&&b<0)
    65         {
    66             x=minn;
    67             y=minn2;
    68             sum=max(a*maxx*maxx+b*minn, a*x*x+b*y);
    69             maxsum=max(sum,a*y*y+b*x);
    70         }
    71         else
    72             maxsum=0;
    73 
    74          printf("Case #%d: %lld
    ",k++,maxsum);
    75     }
    76     return 0;
    77 }
    78 /*HDU 5461*/
  • 相关阅读:
    Linux查看文件被哪个进程占用
    命令行启动rstudio server
    Spring Boot配置文件及多环境配置
    Spring Boot yml配置文件
    js实现自定义概率抽奖算法
    Flutter之adb: failed to install apk的解决方法
    Flutter之不简单的搜索条
    git操作之commit规范
    Flutter之毛玻璃效果的实现
    固定定位下div水平居中
  • 原文地址:https://www.cnblogs.com/weiyuan/p/5761905.html
Copyright © 2011-2022 走看看