交换两数组中的元素使得这两个数组的差最小

交换两数组中的元素使得这两个数组的差最小

a1=A-B

a2=(A-a[i]+b[j])-(B-b[j]=a[i])

=(A-B)-2*(a[i]-b[j])

=a1-2*(a[i]-b[j])

a[i]-b[j]~~(0,a1) min=a1/2;

// ConsoleApplication1.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include<string>using namespace std;void swap(float *A, int lenA, float *B, int lenB, float a1){float min_ = 10000;int k1 = -1;int k2 = -1;for (int i = 0; i < lenA;i++)for (int j = 0; j < lenB; j++){int sub = abs((A[i] - B[j])-a1*1.0/2);if (sub < min_){min_ = sub;k1 = i;k2 = j;}}int temp = A[k1];A[k1] = B[k2];B[k2] = temp;}float FinSum(float A[], int lenA, float B[], int lenB){float Subsum1 = 0;float Subsum2 = 0;for (int i = 0; i < lenA; i++){Subsum1 += A[i];}for (int i = 0; i < lenA; i++){Subsum2 += B[i];}float a1 = Subsum1 - Subsum2;return a1;}int main(){float A[] = { 4,5,6 };int len_a = 3;float B[] = { 1,2,3};int len_b = 3;float a1 = 1000000;float a2 = FinSum(A, len_a, B, len_b);while (a2 < a1){a1 = a2;swap(A, len_a, B, len_b, a2);a2 = FinSum(A, len_a, B, len_b);}return 0;}