題目描述：

????????? 有兩個(gè)數(shù)組a,b，大小都為n,數(shù)組元素的值任意整型數(shù)，無序；
????????? 要求：通過交換a,b中的元素，使數(shù)組a元素的和與數(shù)組b元素的和之間的差最小。?

#include "stdafx.h"#include <math.h> #include <time.h> #include <stdlib.h>void print_arr(int a[], int b[], int n) {printf("a: ");for (int i=0;i<n;i++){printf("%d ", a[i]);}printf("\n");printf("b: ");for (int i=0;i<n;i++){printf("%d ", b[i]);}printf("\n");printf("\n"); }bool swap_min(int a[], int b[], int n) {printf("before swap:\n");print_arr(a, b, n);// 初始化差值int sum_a=0;int sum_b=0;for (int k=0;k<n;k++){sum_a += a[k];sum_b += b[k];}int delta_init = abs(sum_a - sum_b);for (int i=0;i<n;i++){/*if (delta_init == 0) // 優(yōu)化處理{break;}*/int sum_a=0;int sum_b=0;for (int k=0;k<n;k++){sum_a += a[k];sum_b += b[k];}// 假設(shè)交換，差值是否更小bool bswp = false;int jpos = 0;for (int j=0;j<n;j++){int sum_atemp = sum_a-a[i]+b[j];int sum_btemp = sum_b-b[j]+a[i];int delta_temp = abs(sum_atemp-sum_btemp);if (delta_temp < delta_init){delta_init = delta_temp;jpos = j;bswp =true;}}// 假設(shè)成功，執(zhí)行交換元素if (bswp){int tmp = a[i];a[i] = b[jpos];b[jpos] = tmp;}}printf("after swap:\n");print_arr(a,b,n);return true; }int _tmain(int argc, _TCHAR* argv[]) {int a[5] = {0, 0, 0, 0, 9};//1,2,3,4,5int b[5] = {1, 2, 5, 4, 3};//6,7,8,9,100int c[] = {1,2,3,4,5};int d[] = {6,7,8,9,100};int e[]={1,2,3,4,5};int f[]={10,20,30,40,50};swap_min(a, b, 5);swap_min(c, d, 5);swap_min(e, f, 5);errno_t err; unsigned int number; err = rand_s(&number); int m[100];int n[100];srand( (unsigned)time(NULL)); //生成種子for (int i=0;i<100;i++){m[i]=rand()%100;n[i]=abs(rand()%200-100);}swap_min(m, n, 10);getchar();return 0; }
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