题目链接：http://poj.org/problem?id=3399
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 2837 | Accepted: 686 | Special Judge | ||
Description
There is an array of N integer numbers in the interval from -30000 to 30000. The task is to select K elements of this array with maximal possible product.
Input
The input consists of N + 1 lines. The first line contains N and K (1 ≤ K ≤ N ≤ 100) separated by one or several spaces. The others contain values of array elements.
Output
The output contains a single line with values of selected elements separated by one space. These values must be in non-increasing order.
Sample Input
4 2 1 7 2 0
Sample Output
7 2
Source
思路：每次寻找最小的两个负数和最大的两个正数的乘积中较大的；
代码如下：
#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;
#define MAXN 117
int main()
{
int n, k;
int a[MAXN], numz[MAXN], numf[MAXN];
int ans[MAXN];
int i, j;
int num1, num2;
while(~scanf("%d %d",&n,&k))
{
num1 = num2 = 0;
for(i = 0; i < n; i++)
{
scanf("%d",&a[i]);
if(a[i] >= 0)
numz[num1++] = a[i];//大于等于零
else
numf[num2++] = a[i];//负数
}
sort(numz,numz+num1);
sort(numf,numf+num2);
int cont = 0;
if(k&1)
{
k--;
if(num1 > 0)
ans[cont++] = numz[--num1];//k为奇数，且有正数，那么结果中必然会有至少一个正数
else//没有大于等于零的数，即全为负数
{
for(i = num2-1; i > num2-k-1; i--)
{
printf("%d ",numf[i]);
}
printf("%d\n",numf[num2-k-1]);
continue;
}
}
j = 0;
for(i = 0; i < k/2; i++)
{
int t1 = -4017;//初始化为一个小于给定范围的数字
int t2 = -4017;
if(num1 == 1 && num2-j == 1)
{
ans[cont++] = numz[--num1];
ans[cont++] = numf[++j];
}
else
{
if(num1 > 1)
t1 = numz[num1-1]*numz[num1-2];
if(num2-j > 1)
t2 = numf[j] * numf[j+1];
if(t1 > t2)
{
ans[cont++] = numz[--num1];
ans[cont++] = numz[--num1];
}
else
{
ans[cont++] = numf[j++];
ans[cont++] = numf[j++];
}
}
}
sort(ans,ans+cont);
for(i = cont-1; i > 0; i--)//从大到小输出
{
printf("%d ",ans[i]);
}
printf("%d\n",ans[0]);
}
return 0;
}
poj 3399 Product（数学题）,布布扣,bubuko.com
原文地址：http://blog.csdn.net/u012860063/article/details/38365719
