题解 | #牛牛的数组匹配#

#include<stdio.h>
int main(void) 
{
    int n,m;
    int sum1 = 0;
    int sum2 = 0;
    int min = 0;
    scanf("%d %d",&n,&m);
    int arr1[n];
    int arr2[m];
    int i,j;
    int a,b;
    for(i = 0; i < n; i++)
    {
        scanf("%d",&arr1[i]);
        sum1 += arr1[i];//计算数组a的和
    }
    
    min = sum1;

    for(i = 0; i < m; i++)
    {
        scanf("%d",&arr2[i]);    
    } 

    for(int i = 0; i < m; i++)//遍历数组
    {
        sum2 = 0;
        for(int j = i; j < m; j++)//以i为起点，再次遍历
        {
            sum2 += arr2[j];//累加1项，判断一下

            if(abs(sum2-sum1) < min)//取出绝对值最小值
            {
                min = abs(sum2-sum1);//附最小值
                a = i;//记录数组b的坐标
                b = j;
            }
        }
    }

    for(i = a;i<=b;i++)
    {
        printf("%d ",arr2[i]);
    }
    
    return 0;
}