## 一、单项选择题（共15题，每题2分，共计30分）

## 二、阅读程序（判断题1分，选择题3分，共计37分） 判断题正确填 `T`，错误填 `F`。 ### 第1题 ```cpp void recursion(int n) { if (n == 0) return; int lowbit = n % 3; recursion(n / 3); std::cout << lowbit; } void iteration(int n) { int buffer[10]; int size = 0; while (n > 0) { buffer[size++] = n % 3; n = n / 3; } while (size > 0) { std::cout << buffer[--size]; } } ```

### 第2题 ```cpp int solve1(int n) { int s = 0; int f = 1; for (int i = 1; i <= n; ++i) { f = f * i; int m = f; while (m % 10 == 0) { s++; m = m / 10; } } return s; } int solve2(int n) { int t = 0; while (n > 0) { t = t + n / 5; n = n / 5; } return t; } ```

### 第三题 ```cpp #include<iostream> int play[3][3] = {0}; int score[3][3]; bool check_row(int i, int role) { if (play[i][0] != role) return false; if (play[i][1] != role) return false; if (play[i][2] != role) return false; return true; } bool check_col(int j, int role) { if (play[0][j] != role) return false; if (play[1][j] != role) return false; if (play[2][j] != role) return false; return true; } bool check_diag(int role) { if (play[1][1] != role) return false; if (play[0][0] == role && play[2][2] == role) return true; if (play[0][2] == role && play[2][0] == role) return true; return false; } bool check(int i, int j, int role) { return check_row(i, role) || check_col(j, role) || check_diag(role); } int adv(int step) { if (step == 9) return 0; int role = (step % 2) + 1; const int inf = 1000000000; int best = -inf; for (int i = 0; i < 3; i++) { for (int j = 0; j < 3; j++) { if (play[i][j] == 0) { play[i][j] = role; if (check(i, j, role)) { best = inf; } else { int test = score[i][j] - adv(step + 1); if (best < test) { best = test; } } play[i][j] = 0; } } } return best; } int main() { for (int i = 0; i < 3; ++i) for (int j = 0; j < 3; ++j) std::cin >> score[i][j]; int d = adv(0); if (d > 0) std::cout << "First Win\n"; else std::cout << "Second Win\n"; } ```

## 三、完善程序（单选题，每小题3分，共计33分） ### 第1题 有 $n$ 名学生参加了一场考试，分数没有公开。现有 $m$ 条线索，其中第 $i$ 条线索确定第 $x_i$ 号学生的分数高于第 $y_i$ 号学生。又确定没有人分数是相同的。根据这些线索，将学生的分数排序后，多少人的名次是可以确定的呢？保证给定的线索没有矛盾。 ```cpp #include<iostream> int main() { const int maxn = 100; bool f[maxn][maxn] = {false}; int c[maxn] = {0}; int n, m; std::cin >> n >> m; for (int i = 0; i < m; ++i) { int x, y; std::cin >> x >> y; ____(1)____; } for (int _(2)_ = 0; _(2)_ < n; ++ _(2)_) for (int _(3)_ = 0; _(3)_ < n; ++ _(3)_) for (int _(4)_ = 0; _(4)_ < n; ++ _(4)_) if (____(5)____) f[i][j] = true; for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) if (i != j) if (___(6)___) c[i]++; int ans = 0; for (int i = 0; i < n; ++i) if (___(7)___) ans ++; std::cout << ans << '\n'; } ```

### 第2题 给定 $n$ 个整数 $a_0,a_1,\ldots,a_{n-1}$，其中 $1\le n\le40$，请统计，这个序列有多少种子序列，其子序列的和大于 $0$？ ```cpp #include<iostream> #include<algorithm> int a[40]; int x[1 << 20]; int y[1 << 20]; int generate(int begin, int end, int sum, int* out, int pos) { if (begin == end) { ____(1)____; return 1; } else { int f = generate(begin + 1, end, sum + a[begin], out, pos); int s = generate(begin + 1, end, ____(2)____); return f + s; } } int main() { int n; std::cin >> n; for (int i = 0; i < n; ++i) std::cin >> a[i]; int x_size = ____(3)____; int y_size = ____(4)____; std::sort(x, x + x_size); std::sort(y, y + y_size); int j = y_size; long long pair = 0; for (int i = 0; i < x_size; ++i) { while (j > 0 && ____(5)____) { j--; } pair += ____(6)____; } std::cout << pair << "\n"; } ```