Create 1.20.cpp

1 绪论.md

@@ -123,5 +123,89 @@ int main() {
 ### 编写算法，处理上述表格，以统计各院校的男女总分和团体总分，并输出。
 
 ```cpp
+#include <iostream>
+using namespace std;
 
+#define MAXSIZE 100
+
+typedef enum {A, B, C, D, E} SchoolName;
+typedef enum {FEMALE, MALE} Gender;
+
+typedef struct {
+    char event[3];
+    Gender gender;
+    SchoolName school;
+    int score;
+    int point;
+} Component;
+
+typedef struct {
+    int maleSum;
+    int femaleSum;
+    int totalSum;
+} Sum;
+
+for (int i = 0; i < MAXSIZE; ++i) {
+    result[report[i].school].totalSum += report[i].point;
+    switch (report[i].gender) {
+        case 0:
+            result[report[i].school].femaleSum += report[i].point;
+        case 1:
+            result[report[i].school].maleSum += report[i].point;
+    }
+}
+
+for (int i = A; i < E; ++i) {
+    cout << ...
+}
 ```
+
+### 1.19 试编写算法，计算 $i! \cdot 2^i (i = 0, 1, \cdots, n - 1)$ 的值并分别存入数组 `a[arrsize]` 的各个分量中。假设计算机允许的整数最大值为 `MAXINT`，则当 $n > arrsize$ 或对某个 $k (0 \le k \le n - 1)$ 使 $k! \cdot 2^k > MAXINT$ 时，应按出错处理。注意选择你认为较好的出错处理方法。
+
+```cpp
+#include <iostream>
+#include <algorithm>
+#include <cmath>
+using namespace std;
+
+#define ARRSIZE 20
+#define MAXINT 1000000000
+
+int a[ARRSIZE];
+
+int main() {
+    a[0] = 1;
+    cout << a[0];
+    for (int i = 1; a[i - 1] <= MAXINT / 2 / i; ++i) {
+        a[i] = a[i - 1] * i * 2;
+        cout << a[i] << endl;
+    }
+    cout << "OVERFLOW";
+    return 0;
+}
+```
+
+### 1.20 试编写算法求一元多项式 $P_n(x) = \sum\limits_{i=0}^{n}a_ix^i$ 的值 $P_n(x_0)$，并确定算法中每一语句的执行次数和整个算法的时间复杂度。注意选择你认为较好的输入和输出方法。本题的输入为 $a_i (i = 0, 1, \cdots, n)$， $x_0$ 和 $n$，输出为 $P_n(x_0)$。
+
+```cpp
+#include <iostream>
+#include <cmath>
+using namespace std;
+
+int main() {
+    int n, x, p = 0, tmp = 1;
+    cin >> n >> x;
+    int a[n];
+    if (n < 0) {
+        cout << "ERROR" << endl;
+        main();
+    } // O(1)
+    for (int i = 0; i <= n; ++i) {
+        cin >> a[i];
+        p += a[i] * tmp;
+        tmp *= x;
+    } // O(n)
+    cout << p;
+    return 0;
+}
+```

1.18.cpp

@@ -1,8 +1,9 @@
 #include <iostream>
-#include <algorithm>
 using namespace std;
 
-typedef enum {A, B, C, D, E} SchoolNmae;
+#define MAXSIZE 100
+
+typedef enum {A, B, C, D, E} SchoolName;
 typedef enum {FEMALE, MALE} Gender;
 
 typedef struct {
@@ -19,8 +20,16 @@ typedef struct {
     int totalSum;
 } Sum;
 
-int main() {
-    while 
-    cin >> item[i].name >> item[i].gender >> item[i].score >> item[i].point;
-
+for (int i = 0; i < MAXSIZE; ++i) {
+    result[report[i].school].totalSum += report[i].point;
+    switch (report[i].gender) {
+        case 0:
+            result[report[i].school].femaleSum += report[i].point;
+        case 1:
+            result[report[i].school].maleSum += report[i].point;
+    }
+}
+
+for (int i = A; i < E; ++i) {
+    cout << ...
 }

1.19.cpp

@@ -0,0 +1,22 @@
+// 试编写算法，计算 $i! \cdot 2^i (i = 0, 1, \cdots, n - 1)$ 的值并分别存入数组 `a[arrsize]` 的各个分量中。假设计算机允许的整数最大值为 `MAXINT`，则当 $n > arrsize$ 或对某个 $k (0 \le k \le n - 1)$ 使 $k! \cdot 2^k > MAXINT$ 时，应按出错处理。注意选择你认为较好的出错处理方法。
+
+#include <iostream>
+#include <algorithm>
+#include <cmath>
+using namespace std;
+
+#define ARRSIZE 20
+#define MAXINT 1000000000
+
+int a[ARRSIZE];
+
+int main() {
+    a[0] = 1;
+    cout << a[0];
+    for (int i = 1; a[i - 1] <= MAXINT / 2 / i; ++i) {
+        a[i] = a[i - 1] * i * 2;
+        cout << a[i] << endl;
+    }
+    cout << "OVERFLOW";
+    return 0;
+}

1.20.cpp

@@ -0,0 +1,22 @@
+// 试编写算法求一元多项式 $P_n(x) = \sum\limits_{i=0}^{n}a_ix^i$ 的值 $P_n(x_0)$，并确定算法中每一语句的执行次数和整个算法的时间复杂度。注意选择你认为较好的输入和输出方法。本题的输入为 $a_i (i = 0, 1, \cdots, n)$， $x_0$ 和 $n$，输出为 $P_n(x_0)$。
+
+#include <iostream>
+#include <cmath>
+using namespace std;
+
+int main() {
+    int n, x, p = 0, tmp = 1;
+    cin >> n >> x;
+    int a[n];
+    if (n < 0) {
+        cout << "ERROR" << endl;
+        main();
+    } // O(1)
+    for (int i = 0; i <= n; ++i) {
+        cin >> a[i];
+        p += a[i] * tmp;
+        tmp *= x;
+    } // O(n)
+    cout << p;
+    return 0;
+}