题目地址:点击打开链接
题意:求ax^2+by的最大值,x和y在一个数组中
思路:2层暴力超时,那就一层暴力ax^2的所有情况,参考学长A的,思路不是很清晰,等我问清楚了再写详细的
AC代码:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
using namespace std;
const int maxn = 5e6 + 10;
long long map1[maxn];
int main()
{
int t,i;
int n,a,b,cas=1;
scanf("%d",&t);
while(t--)
{
long long ans = -9223372036854775;
long long temp;
scanf("%d%d%d",&n,&a,&b);
for(i=1; i<=n; i++)
{
scanf("%I64d",&map1[i]);
}
if(a == 0 && b == 0)
{
int l = 0;
printf("Case #%d: %d\n",cas++,l);
continue;
}
sort(map1+1,map1+n+1);
for(i=1; i<=n; i++)
{
temp = a * map1[i] * map1[i];
if(i != 1 && i != n)
{
temp += b * map1[1] > b * map1[n] ? b * map1[1] : b * map1[n];
}
else if(i == n)
{
temp += b * map1[n-1] > b * map1[1] ? b * map1[n-1] : b * map1[1];
}
else if(i == 1)
{
temp += b * map1[n] > b * map1[2] ? b * map1[n] : b * map1[2];
}
if(temp > ans)
ans = temp;
}
printf("Case #%d: %I64d\n",cas++,ans);
}
return 0;
}