普及组 CSP-J 2026 初赛模拟卷 5

（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 | using namespace std;
3 |
4 | int process(int n) {
5 |     if (n == 0) return 0;
6 |     return n % 2 + process(n / 2) * 10;
7 | }
8 |
9 | int main() {
10 |     int num;
11 |     cin >> num;
12 |     int result = process(num);
13 |     cout << result << endl;
14 |     return 0;
15 | }

1 | #include <iostream>
2 | #include <cmath>
3 | #define MAX 100
4 | using namespace std;
5 | int digits[MAX + 9];
6 | void solve(int a, int b, int precision) {
7 |     if (b == 0) {
8 |         cout << "Error: Division by zero!" << endl;
9 |         return;
10 |     }
11 |     if (precision > MAX) precision = MAX;
12 |     if (precision < 0) precision = 0;
13 |
14 |     bool negative = (a < 0 && b > 0) || (a > 0 && b < 0);
15 |     a = abs(a);
16 |     b = abs(b);
17 |     digits[0] = a / b;
18 |     int remainder = a % b;
19 |
20 |     for (int i = 1; i <= precision + 1; i++) {
21 |         remainder *= 10;
22 |         digits[i] = remainder / b;
23 |         remainder %= b;
24 |     }
25 |
26 |     if (digits[precision + 1] >= 5) {
27 |         digits[precision]++;
28 |         for (int i = precision; i >= 1; i--) {
29 |             if (digits[i] >= 10) {
30 |                 digits[i] -= 10;
31 |                 digits[i - 1]++;
32 |             } else {
33 |                 break;
34 |             }
35 |         }
36 |     }
37 |
38 |     if (negative) cout << "-";
39 |     cout << digits[0] << '.';
40 |     for (int i = 1; i <= precision; i++) {
41 |         cout << digits[i];
42 |     }
43 |     cout << endl;
44 | }
45 |
46 | int main() {
47 |     int a, b, prec;
48 |     cin >> a >> b >> prec;
49 |     solve(a, b, prec);
50 |     return 0;
51 | }
52 | // 所有输入数据的绝对值均小于 1000

1 | #include <iostream>
2 | #include <vector>
3 | using namespace std;
4 |
5 | int bSearch(vector<int>& a, int target) {
6 |     int left = 0, right = a.size() - 1;
7 |
8 |     while (left <= right) {
9 |         int mid = left + (right - left) / 2;
10 |         if (a[mid] == target) {
11 |             return mid;
12 |         } else if (a[mid] > target) {
13 |             left = mid + 1;
14 |         } else {
15 |             right = mid - 1;
16 |         }
17 |     }
18 |     return -1;
19 | }
20 |
21 | int main() {
22 |     int n, x;
23 |     cin >> n >> x;
24 |
25 |     vector<int> a(n);
26 |     for (int i = 0; i < n; i++) cin >> a[i];
27 |     int ans = bSearch(a, x);
28 |     cout << (ans != -1 ? ans : -1) << endl;
29 |     return 0;
30 | }

给定一个整数数组和一个整数 $k$，程序找出数组中 $k$ 个不相邻元素（即任意两个被选元素在数组中不相邻）的最大和。

1 | #include <iostream>
2 | #include <vector>
3 | #include <algorithm>
4 | using namespace std;
5 |
6 | int solve(vector<int>& nums, int k) {
7 |     int n = ①;
8 |     vector<vector<int>> dp(②);
9 |
10 |     for (int i = 1; i <= n; i++) {
11 |         for (int j = 1; j <= k; j++) {
12 |             if (j == 1) {
13 |                 dp[i][j] = max(③);
14 |             } else {
15 |                 dp[i][j] = max(dp[i - 1][j], ④);
16 |             }
17 |         }
18 |     }
19 |     return ⑤;
20 | }
21 |
22 | int main() {
23 |     int n, k;
24 |     cin >> n >> k;
25 |     vector<int> nums(n);
26 |     for (int i = 0; i < n; i++) cin >> nums[i];
27 |     cout << solve(nums, k) << endl;
28 |     return 0;
29 | }

给定一条长度为 $L$ 的线段（$L$ 为整数），以及 $n$ 个整数区间，每个区间有整数起始位置和结束位置。程序的目标是选择最少数量的区间，使得这些区间完全连续覆盖整个线段 $[0,L]$。如果无法完全覆盖，则输出 -1。

1 | #include <iostream>
2 | #include <algorithm>
3 | using namespace std;
4 | const int N = 100005;
5 |
6 | struct Interval { int s, e; } ivl[N];
7 | bool cmp(Interval a, Interval b) {
8 |     return a.s < b.s || (①);
9 | }
10 |
11 | int main() {
12 |     int L, n;
13 |     cin >> L >> n;
14 |
15 |     for (int i = 0; i < n; i++) {
16 |         cin >> ivl[i].s >> ivl[i].e;
17 |     }
18 |     ②
19 |
20 |     int curr = 0;
21 |     int cnt = 0;
22 |     int i = 0;
23 |
24 |     while (curr < L) {
25 |         int mx_reach = curr;
26 |         while (③) {
27 |             if (ivl[i].e > mx_reach) {
28 |                 mx_reach = ivl[i].e;
29 |             }
30 |             i++;
31 |         }
32 |
33 |         if (④) {
34 |             cout << -1 << endl;
35 |             return 0;
36 |         }
37 |         cnt++;
38 |         ⑤
39 |     }
40 |
41 |     cout << cnt << endl;
42 |     return 0;
43 | }