普及组 CSP-J 2026 初赛模拟卷 2

（共 15 题，每题 2 分，共计 30 分；每题有且仅有一个正确选项）

程序输入不超过数组或字符串定义的范围；判断题正确填 ✓，错误填 ✗；除特殊说明外，判断题每题 2 分，选择题每题 3 分。

 1 | #include <iostream>
 2 | using namespace std;
 3 | void hanoi(int n, char a, char b, char c) {
 4 |     if (n == 1)
 5 |         cout << n << "_" << a << "_" << c << endl;
 6 |     else {
 7 |         hanoi(n - 1, a, c, b);
 8 |         cout << n << "_" << a << "_" << c << endl;
 9 |         hanoi(n - 1, b, a, c);
10 |     }
11 | }
12 | int main() {
13 |     int n;
14 |     cin >> n;
15 |     hanoi(n, 'A', 'B', 'C');
16 |     return 0;
17 | }

 1 | #include <cstdio>
 2 | #define N 1005
 3 | using namespace std;
 4 | int num[N];
 5 | int main() {
 6 |     int a1 = 1, n, x;
 7 |     scanf("%d", &n);
 8 |     num[1] = 1;
 9 |     for (int i = 1; i <= n; ++i) {
10 |         x = 0;
11 |         for (int j = 1; j <= a1; ++j) {
12 |             num[j] = num[j] * 5 + x;
13 |             x = num[j] / 10;
14 |             num[j] %= 10;
15 |         }
16 |         if (x > 0) num[++a1] = x;
17 |     }
18 |     printf("0.");
19 |     for (int i = a1; i < n; ++i) putchar('0');
20 |     for (int i = a1; i >= 1; i--) printf("%d", num[i]);
21 |     putchar('\n');
22 |     return 0;
23 | }

 1 | #include <iostream>
 2 | using namespace std;
 3 | int l, n, m, a[50005], ans;
 4 | bool check(int dis) {
 5 |     int count = 0, last = 0;
 6 |     for (int i = 1; i <= n + 1; i++)
 7 |         if (a[i] - last < dis) count++;
 8 |         else last = a[i];
 9 |     if (count > m) return 0;
10 |     return 1;
11 | }
12 |
13 | int main() {
14 |     // 输入保证 l, n, m, a[i] 是正整数，且 a[i] 严格递增
15 |     cin >> l >> n >> m;
16 |     for (int i = 1; i <= n; i++) cin >> a[i];
17 |     a[n + 1] = l;
18 |     int fl = 0, fr = l;
19 |     while (fl <= fr) {
20 |         int mid = (fl + fr) / 2;
21 |         if (check(mid)) fl = mid + 1, ans = mid;
22 |         else fr = mid - 1;
23 |     }
24 |     cout << ans << endl;
25 |     return 0;
26 | }

（无权图最短路）给定一个 $n$ 个顶点、$m$ 条边的无向图以及起点 $s$，使用 BFS 求起点到每个顶点的最少边数。不可达顶点的距离保持为 -1。

1 | #include <iostream>
2 | #include <queue>
3 | #include <vector>
4 | using namespace std;
5 | int main() {
6 |     int n, m, s;
7 |     cin >> n >> m >> s;
8 |     vector<vector<int>> g(n + 1);
9 |     for (int i = 0; i < m; ++i) {
10 |         int u, v;
11 |         cin >> u >> v;
12 |         g[u].push_back(v);
13 |         g[v].push_back(u);
14 |     }
15 |     vector<int> dist(n + 1, -1);
16 |     queue<int> q;
17 |     ① = 0;
18 |     ②;
19 |     while (!q.empty()) {
20 |         int u = q.front();
21 |         ③;
22 |         for (int v : g[u]) {
23 |             if (④) {
24 |                 ⑤;
25 |                 q.push(v);
26 |             }
27 |         }
28 |     }
29 |     for (int i = 1; i <= n; ++i)
30 |         cout << dist[i] << (i == n ? '\n' : ' ');
31 |     return 0;
32 | }

（0-1 背包）有 $n$ 件物品，背包容量为 $V$。第 $i$ 件物品重量为 $w_i$、价值为 $c_i$，每件物品最多选择一次，求不超过容量时的最大总价值。

1 | #include <algorithm>
2 | #include <iostream>
3 | #include <vector>
4 | using namespace std;
5 | int main() {
6 |     int n, V;
7 |     cin >> n >> V;
8 |     vector<int> dp(V + 1, ①);
9 |     for (int i = 1; i <= ②; ++i) {
10 |         int w, c;
11 |         cin >> w >> c;
12 |         for (int j = V; ③; --j)
13 |             dp[j] = ④;
14 |     }
15 |     cout << ⑤ << '\n';
16 |     return 0;
17 | }