POJ 2209 The King(简单贪心)
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 7499 | Accepted: 4060 |
Description
Nevertheless, he did not consider this to be a really great drawback, since he had a lot of wizards who could count up to one hundred (and some of them, people said, even up to one thousand), so it was all right. But one day the grief came to the kingdom as the outnumbering barbarians started to approach from all sides. And the king then had to make the most important decision in his life. He had to choose which of his sons to make generals that he would send to the borders of the country to lead the army.
However, the king knew that though some of his sons were clever, just like he was, some of them were quite stupid and could only lower army spirits with their wrong decisions. More precisely, he knew about each of his sons his mental potential -- an integer number ranging from minus three to three (remember, that the king could count only up to three). He also knew that the chance of his army defeating barbarians was proportional to the sum of some powers of mental potentials of those of his sons that he would make generals (the power exponent was a positive integer number, the same for all his sons and not exceeding three either). Thus he had to choose such a combination of his sons to lead
the army, that this sum would be maximal possible.
However, the king himself could not make all apropriate calculations since, for example, the second power of the number not exceeding three (which is its square) could be greater than three, and therefore he asked you, his most intellegent wizard, to solve this problem.
Input
Output
Sample Input
3
3
2 -1 1
Sample Output
9
Hint
Source
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
int n,m;
int a[];
int b[];
while(scanf("%d",&n)!=EOF)
{
scanf("%d",&m);
for(int i=;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
int sum=,ans=;
for(int i=;i<n;i++)
{
for(int j=;j<=m;j++)
sum*=a[i];
if(sum>)
ans+=sum;
sum=;
}
printf("%d\n",ans);
}
return ;
}
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