交換兩序列a,b中的元素，使序列a的和與序列b的和之間的差值最小

參考網址：http://blog.csdn.net/v_JULY_v/article/details/6126444 第32題

問題描述：

有兩個序列a,b，大小都為n,序列元素的值任意整數，無序；
要求：通過交換a,b中的元素，使[序列a元素的和]與[序列b元素的和]之間的差最小

代碼實現：

#include <cstdio>#include <cstdlib>#include <cmath>void swapi(int *x, int *y);int sum(int a[], int n);int isXinA(int x, int A);void process(int a[], int b[], int n);int main(){int a[] = {1, 2, 3, 29, 30};int b[] = {1, 210, 232, 12311, 12312};int n = 5;process(a, b, n);for (int i = 0; i < n; i++)printf("%d  ", a[i]);printf("\n");for (int i = 0; i < n; i++)printf("%d  ", b[i]);printf("\n%d\n", abs(sum(a, n)- sum(b, n)));system("pause");return 0;}void swapi(int *x, int *y){int temp = *x;*x = *y;*y = temp;}int sum(int a[], int n){int ret = 0;for (int i = 0; i < n; i++)ret += a[i];return ret;}int isXinA(int x, int A){if ((x>0 && x<A) || (x>A && x<0))return 0;return 1;}void process(int a[], int b[], int n){int sum_a = sum(a, n);int sum_b = sum(b, n);int A = sum_a - sum_b;float minValue = abs(a[0]-b[0] - A/2.0);int ii = 0, jj = 0;int flag = 0;if (A == 0)return;for (int i = 0; i < n; i++){for (int j = 0; j < n; j++){int x = a[i] - b[j];if (x == 0)continue;if (isXinA(x, A) == 0){float temp = (float)abs(a[i]-b[j] - A/2.0);if (temp < minValue){minValue = temp;ii = i;jj = j;flag = 1;}}}}if (flag == 1){swapi(&a[ii], &b[jj]);process(a, b, n);//繼續求解}}//求解思路：//當前數組a和數組b的和之差為//A = sum(a) - sum(b)////a的第i個元素和b的第j個元素交換後，a和b的和之差為//A' = sum(a) - a[i] + b[j] - （sum(b) - b[j] + a[i])// = sum(a) - sum(b) - 2 (a[i] - b[j])//   = A - 2 (a[i] - b[j])////設x = a[i] - b[j]//|A| - |A'| = |A| - |A-2x|////假設A > 0,//當x 在 (0,A)之間時，做這樣的交換才能使得交換後的a和b的和之差變小，//x越接近A/2效果越好,//如果找不到在(0,A)之間的x，則當前的a和b就是答案。////所以演算法大概如下：//在a和b中尋找使得x在(0,A)之間並且最接近A/2的i和j，交換相應的i和j元素，//重新計算A後，重複前面的步驟直至找不到(0,A)之間的x為止。