题解 | #牛牛的数组匹配#

#include<iostream>
#include<stdlib.h>
#include<math.h>
using namespace std;
    //        存放子数组的和，以及左右下标
struct person{
    int left,  right;
    int sum_b = 0;
};
int main()
{    
    int* a, * b;
    //    创建结构体变量
    person sum[2000];
    int n, m, i, j;
    int sum_a = 0, d = 0;
    cin >> n >> m;
    a = (int*)malloc(n * 4);
    b = (int*)malloc(m * 4);
    for (i = 1; i <= m; i++)
        d += i;
    for (i = 0; i < n; i++)
    {
        cin >> a[i];
        sum_a += a[i];
    }
    for (i = 0; i < m; i++)
        cin >> b[i];
    //        初始化
    sum[0].left = 0, sum[0].right = 0;
    //        求和
    for (i = 0; i < d; i++)
    {
        for (j = sum[i].left; j <= sum[i].right; j++)
            sum[i].sum_b += b[j];
        sum[i].sum_b = abs(sum[i].sum_b - sum_a);
            //    左右下标判断
            if (sum[i].right == m - 1)
            {
                sum[i + 1].left = sum[i].left + 1;
                sum[i+1].right = sum[i + 1].left;
            }
            else
            {
                sum[i + 1].left = sum[i].left;
                sum[i+1].right=sum[i].right+1;
            }
    }
        //        判断输出
    sum[d].left = sum[0].left;
    sum[d].right = sum[0].right;
    sum[d].sum_b = sum[0].sum_b;
    for(i=1;i<d;i++)
        if (sum[d].sum_b > sum[i].sum_b)
        {
            sum[d].left = sum[i].left;
            sum[d].right = sum[i].right;
            sum[d].sum_b = sum[i].sum_b;
        }
    for (i = sum[d].left; i <= sum[d].right; i++)
        cout << b[i] << ' ';
    cout << endl;
    system("pause");
    return 0;
}