hdu 5461(2015沈阳网赛 简单暴力) Largest Point

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

题意就是在数组中找出a*t[i]*t[i]+b*t[j]的最大值,特别注意的是这里i和i不能相等,想不到的是居然直接暴力排序都能过,而且这题的数据很水,好多有bug的代码都能过

但是我们要对自己要求严一点,尽量吧各种情况考虑到位,如果下面的代码还有不妥,欢迎指正

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
ll max(ll x,ll y) {
    if (x>y) return x;
    else return y;
}
ll a[5000001];
int main()
{
    ll t,q,i,n,c,w,b,num1,num2;
    while (~scanf("%I64d",&t))
    {
        ll ans=1;
        while (t--){
        scanf("%I64d %I64d %I64d",&n,&c,&b);
        q=100000000;
        for (i=1;i<=n;i++)
            scanf("%I64d",&a[i]);
        sort(a+1,a+n+1);
        for (i=1;i<=n;i++)
            if (fabs(a[i])<q) q=fabs(a[i]),w=i;
        printf("Case #%I64d: ",ans++);
        if (c<0&&b<0)
        {
            if (w!=1) printf("%I64d
",c*a[w]*a[w]+b*a[1]);
            else
            {
                num1=c*a[1]*a[1]+b*a[2];
                num2=c*a[2]*a[2]+b*a[1];
                printf("%d
",max(num1,num2));
            }
            continue;
        }
        else if (c>0&&b>0)
        {
            printf ("%I64d
",max(max(c*a[n]*a[n]+b*a[n-1], c*a[n-1]*a[n-1]+b*a[n]), c*a[1]*a[1]+b*a[n]));
            continue;
        }
        else if (c>=0&&b<=0)
        {
            printf ("%I64d
",max(max(c*a[n]*a[n]+b*a[1], c*a[1]*a[1]+b*a[2]), c*a[2]*a[2]+b*a[1]));
            continue;
        }
        else if (c<=0&&b>=0)
        {
            if (w!=n) printf("%I64d
",c*a[w]*a[w]+b*a[n]);
            else
            {
                num1=c*a[n]*a[n]+b*a[n-1];
                num2=c*a[n-1]*a[n-1]+b*a[n];
                printf("%d
",max(num1,num2));
            }
            continue;
        }
        }
    }
    return 0;
}