题解 | 牛牛的数组匹配

#include <stdio.h>
#include <stdlib.h>   // for abs()

int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    int a[100] = {0}, b[100] = {0};
    int sum1 = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
        sum1 += a[i];
    }
    for (int i = 0; i < m; i++) {
        scanf("%d", &b[i]);
    }

    int min = 1e9;      // 足够大的初值
    int k = 0, l = 0;   // 记录最优子数组的区间 [k, l)

    for (int i = 0; i < m; i++) {
        int sum2 = 0;
        for (int j = i; j < m; j++) {
            sum2 += b[j];               // 当前子数组 b[i]~b[j]
            int diff = abs(sum1 - sum2);
            if (diff < min) {
                min = diff;
                k = i;
                l = j + 1;              // 结束下标+1，便于后续输出
            }
        }
    }

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