题解 | #牛牛的数组匹配#

#include <stdio.h>
#include<stdlib.h>

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    int ar1[100], ar2[100];
    int i = 0, j = 0, sum1 = 0, sum2 = 0;
    for (i = 0; i < n; i++)
    {
        scanf("%d", &ar1[i]);
        sum1 += ar1[i];
    }
    for (i = 0; i < m; i++)
    {
        scanf("%d", &ar2[i]);
        sum2 += ar2[i];
    }
    int min = 1000, x=0, y=0;//x为左边界  y为右边界
    for (i = 0; i < m; i++)//起始位置
    {
        int sum=sum2,t=0;
        for (t = 0; t < i; t++)
        {
            sum -= ar2[t];
        }
        for (j = m - 1; j > i; j--)//从后向前
        {
            if (min > abs(sum - sum1))
            {
                min = abs(sum - sum1);
                x = i;
                y = j;
            }
            sum -= ar2[j];
        }
    }
    int p = 0;
    for (p = x; p <= y; p++)
    {
        printf("%d ", ar2[p]);
    }
    return 0;
}