- #include <algorithm>
- #include <iostream>
- using namespace std;
- int n;
- int d[50][2];
- int ans;
- void dfs(int n, int sum) {
- if (n == 1) {
- ans = max(sum, ans);
- return;
- }
- for (int i = 1; i < n; ++i) {
- int a = d[i - 1][0], b = d[i - 1][1];
- int x = d[i][0], y = d[i][1];
- d[i - 1][0] = a + x;
- d[i - 1][1] = b + y;
- for (int j = i; j < n - 1; ++j)
- d[j][0] = d[j + 1][0], d[j][1] = d[j + 1][1];
- int s = a + x + abs(b - y);
- dfs(n - 1, sum + s);
- for (int j = n - 1; j > i; --j)
- d[j][0] = d[j - 1][0], d[j][1] = d[j - 1][1];
- d[i - 1][0] = a, d[i - 1][1] = b;
- d[i][0] = x, d[i][1] = y;
- }
- }
- int main() {
- cin >> n;
- for (int i = 0; i < n; ++i)
- cin >> d[i][0];
- for (int i = 0; i < n; ++i)
- cin >> d[i][1];
- ans = 0;
- dfs(n, 0);
- cout << ans << endl;
- return 0;
- }
假设输入的 n 是不超过 50 的正整数,d[i][0]、d[i][1] 都是不超过 10000 的正整数。
2)判断:若输入的 n 为 20,接下来的输入全为 0,则输出为 0。( )