普及组 CSP-J 2026 初赛模拟卷 6

（共 15 题，每题 2 分，共计 30 分；每题有且仅有一个正确选项）

1 | #include <iostream>
2 | #include <vector>
3 | using namespace std;
4 |
5 | int solve(int n) {
6 |     int sum = 0;
7 |     for (int i = 1; i <= n / 2; i++)
8 |         if (n % i == 0) sum += i;
9 |     return sum;
10 | }
11 |
12 | int main() {
13 |     int n;
14 |     cin >> n;
15 |     int sum = solve(n);
16 |     if (sum == n) {
17 |         cout << "perfect_number" << endl;
18 |     } else if (sum > n) {
19 |         cout << "abundant_number" << endl;
20 |     } else {
21 |         cout << "deficient_number" << endl;
22 |     }
23 |     return 0;
24 | }

1 | #include <bits/stdc++.h>
2 | using namespace std;
3 | const int N = 10009;
4 | int n, nums[N];
5 |
6 | int solve() {
7 |     int res = n;
8 |     for (int i = 0; i < n; i++) {
9 |         res ^= i;
10 |         res ^= nums[i];
11 |     }
12 |     return res;
13 | }
14 |
15 | int main() {
16 |     cin >> n;
17 |     // 保证输入的 n 个数在 0..n 范围内，且互不相同
18 |     for (int i = 0; i < n; i++) cin >> nums[i];
19 |     cout << solve() << endl;
20 |     return 0;
21 | }

1 | #include <bits/stdc++.h>
2 | using namespace std;
3 | const int N = 100009;
4 | const int INF = 1e8 - 1;
5 | int a[N];
6 |
7 | int main() {
8 |     int n, x;
9 |     cin >> n >> x;
10 |     for (int i = 0; i < n; i++) cin >> a[i];
11 |     sort(a, a + n, less<int>());
12 |
13 |     int left = 0, right = n - 1, p = n;
14 |     while (left <= right) {
15 |         int mid = (left + right) / 2;
16 |         if (a[mid] > x) {
17 |             p = mid;
18 |             right = mid - 1;
19 |         } else {
20 |             left = mid + 1;
21 |         }
22 |     }
23 |     cout << p << " ";
24 |     if (p < n) cout << a[p] << endl;
25 |     else cout << INF << endl;
26 |     return 0;
27 | }

（二叉树层序遍历）给定一棵以结点 1 为根的二叉树。结点编号为 $1ldots n$，输入每个结点的左右孩子编号，0 表示对应孩子不存在。请按层序输出所有结点，并统计叶结点数量。保证 $nge1$，输入构成一棵合法二叉树。

1 | #include <iostream>
2 | #include <queue>
3 | #include <vector>
4 | using namespace std;
5 | int main() {
6 |     int n;
7 |     cin >> n;
8 |     vector<int> leftChild(n + 1), rightChild(n + 1);
9 |     for (int i = 1; i <= n; ++i)
10 |         cin >> leftChild[i] >> rightChild[i];
11 |     queue<int> q;
12 |     ①;
13 |     int leaves = 0;
14 |     bool first = true;
15 |     while (!q.empty()) {
16 |         int u = q.front();
17 |         q.pop();
18 |         if (!first) cout << ' ';
19 |         cout << u;
20 |         first = false;
21 |         if (②) q.push(leftChild[u]);
22 |         if (rightChild[u] != 0) ③;
23 |         if (④) ++leaves;
24 |     }
25 |     cout << '\n' << ⑤ << '\n';
26 |     return 0;
27 | }

（网格路径计数）给定一个 $n\times m$ 的网格，其中 `#` 表示障碍，`.` 表示可通行。只能从 $(1,1)$ 向下或向右移动，求到达 $(n,m)$ 的路径数。保证起点和终点可通行，答案不超过 `long long` 范围。

1 | #include <iostream>
2 | #include <string>
3 | using namespace std;
4 | long long dp[1005][1005];
5 | int main() {
6 |     int n, m;
7 |     cin >> n >> m;
8 |     string g[1005];
9 |     for (int i = 1; i <= n; ++i) cin >> g[i];
10 |     dp[1][1] = ①;
11 |     for (int i = 1; ②; ++i) {
12 |         for (int j = 1; j <= m; ++j) {
13 |             if (i == 1 && j == 1) continue;
14 |             if (③) dp[i][j] = 0;
15 |             else dp[i][j] = ④;
16 |         }
17 |     }
18 |     cout << ⑤ << '\n';
19 |     return 0;
20 | }