最小化C ++中由N个数形成的N / 2对的和的平方和
问题陈述
给定n个元素的数组。任务是以这样的方式创建n/2对，即n/2对的平方和最小。
示例
如果给定数组为-
arr[] = {5, 10, 7, 4}
then minimum sum of squares is 340 if we create pairs as (4, 10) and ( 5, 7)算法
1. Sort the array
2. Take two variables which point to start and end index of an array
3. Calulate sum as follows:
sum = arr[start] + arr[end];
sum = sum * sum;
4. Repeate this procedure till start < end and increment minSum as follows:
While (start < end) {
sum = arr[start] + arr[end];
sum = sum * sum;
minSum += sum;
++start;
--end;
}示例
#include <iostream>
#include <algorithm>
#define SIZE(arr) (sizeof(arr) / sizeof(arr[0]))
using namespace std;
int getMinSquareSum(int *arr, int n) {
sort(arr, arr + n);
int minSum = 0;
int start = 0;
int end = n - 1;
while (start < end) {
int sum = arr[start] + arr[end];
sum *= sum;
minSum += sum;
++start;
--end;
}
return minSum;
}
int main() {
int arr[] = {5, 10, 7, 4};
int res = getMinSquareSum(arr, SIZE(arr));
cout << "Minimum square sum: " << res << "\n";
return 0;
}输出结果
当您编译并执行上述程序时。它产生以下输出-
Minimum square sum: 340
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