第8讲 函数 单元作业

 

1. 编一判断m是否为素数的函数，并在主函数中利用它输出十对最小的孪生素数。所谓孪生素数是指两个相差为2的素数，如3和5，11和13。程序运行结果见下图。

函数形式为：

bool isprime(int m)

yzy.ver:

 1 #include "iostream"
 2 using namespace std;
 3 bool isprime(int m)
 4 {
 5     int i;
 6     bool flag = 0;
 7     for (i = 2; ; i++)
 8         if (m % i == 0)
 9             break;
10     if (i == m)
11         flag= 1;
12     return flag;
13 }
14 int main()
15 {
16     int m=2, s = 0;
17     while (s < 10)
18     {
19         if (isprime(m) && isprime(m + 2))
20         {
21             cout << "<" << m << "," << m + 2 << ">" << endl;
22             s++;
23         }
24         m++;
25     }
26     system("pause");
27     return 0;
28 }
View Code

 

cm's best ver:

 1 #include "iostream"
 2 
 3 using namespace std;
 4 
 5 
 6 
 7 bool isprime(int m);
 8 
 9 int main()
10 
11 {
12 
13     int two = 2, count = 0;
14 
15     for(int j = 2; count<10; ++j){
16 
17         if(isprime(j) && isprime(j+two)){
18 
19             count++;
20 
21             printf("(%d,%d)\n",j,j+two);
22 
23         }
24 
25     }
26 
27     return 0;
28 
29 }
30 
31 bool isprime(int m){
32 
33     for(int i=2; i<m; ++i){
34 
35         if(m%i==0)
36 
37             return false;
38 
39     }
40 
41     return true;
42 
43 }
View Code

 

2. 

编一函数，功能为判断一字符串是否为回文，如果是回文则返回1，否则返回0。回文是指顺读和倒读都一样的字符串，如“deed”和“level”是回文。在主函数中对输入的字符串加以调用。

函数形式为：int huiwen(char s[])

yzy.ver:

 1 #include "iostream"
 2 #define N 100
 3 using namespace std;
 4 int huiwen(char s[])
 5 {
 6     int i,m;
 7     m = strlen(s);
 8     bool flag = 1;
 9     for (i = 0; i <= (m - 1) / 2; i++)
10     {
11         if (s[i] != s[m - 1 - i])
12         {
13             flag = 0;
14         }
15     }
16     return flag;
17 }
18 int main()
19 {
20     char s[N];
21     gets_s(s);
22     int flag;
23     flag = huiwen(s);
24     cout << flag << endl;
25     system("pause");
26     return 0;
27 }
View Code

 

cm's best ver:

 1 #include <iostream>
 2 
 3 #include <string.h>
 4 
 5 using namespace std;
 6 
 7 
 8 
 9 int huiwen(char s[]);
10 
11 int main()
12 
13 {
14 
15     const int N = 100;
16 
17     char s[N] = "";
18 
19     cin>>s;
20 
21     if(huiwen(s))
22 
23         cout<<"是回文"<<endl;
24 
25     else
26 
27         cout<<"不是回文"<<endl;
28 
29     return 0;
30 
31 }
32 
33 int huiwen(char s[])
34 
35 {
36 
37     char *p = s, *q = s+strlen(s)-1;
38 
39     while(p<q)
40 
41     {
42 
43         if(*p!=*q)
44 
45             return 0;
46 
47         p++, q--;
48 
49     }
50 
51     return 1;
52 
53 }
View Code

 

3. 

函数的功能是将学生成绩从高分到低分排序，并统计优秀与不及格的人数。用下面两种方法实现：

（1）函数形式为：int fun1(int s[],int n,int *x)

要求优秀人数通过return返回，不及格人数通过指针参数返回结果。

（2）函数形式为：void fun2(int s[],int n,int &x,int &y)

要求优秀与不及格的人数通过引用参数返回结果。

分别编二个函数，学生人数从键盘输入。

yzy.ver:

  1 //函数一：
  2 #include "iostream"
  3 #define N 100
  4 using namespace std;
  5 int fun1(int s[], int n, int *x)
  6 {
  7     int i,j,k,t,r=0,p=0;
  8     for (i = 0; i < n-1 ; i++)
  9     {
 10         k = i;
 11         for (j = i + 1; j < n; j++)
 12             if (s[k] < s[j])
 13                 k = j;
 14         if (i != k)
 15         {
 16             t = s[i];
 17             s[i] = s[k];
 18             s[k] = t;
 19         }
 20     }
 21     for (i = 0; i < n; i++)
 22     {
 23         if (s[i] >= 90)
 24             r++;
 25         if (s[i] < 60)
 26             p++;
 27     }
 28     *x = p;
 29     return r;
 30 }
 31 int main()
 32 {
 33     int s[N];
 34     cout << "输入学生人数：";
 35     int n,i, num=0;
 36     cin >> n;
 37     cout << "输入学生成绩（满分100）：";
 38     for (i = 0; i < n; i++)
 39     {
 40         cin >> s[i];
 41     }
 42     fun1(s, n, &num);
 43     cout << "优秀人数为" << fun1(s, n, &num) << endl;
 44     cout << "不及格人数为" << num << endl;
 45     cout << "从高到低排序成绩：" << endl;
 46     for (i = 0; i < n; i++)
 47     {
 48         cout << s[i]<<' ';
 49     }
 50     system("pause");
 51     return 0;
 52 }
 53 //函数二：
 54 #include "iostream"
 55 #define N 100
 56 using namespace std;
 57 void fun2(int s[], int n, int& x, int& y)
 58 {
 59     int i, j, k, t, r = 0, p = 0;
 60     for (i = 0; i < n - 1; i++)
 61     {
 62         k = i;
 63         for (j = i + 1; j < n; j++)
 64             if (s[k] < s[j])
 65                 k = j;
 66         if (i != k)
 67         {
 68             t = s[i];
 69             s[i] = s[k];
 70             s[k] = t;
 71         }
 72     }
 73     for (i = 0; i < n; i++)
 74     {
 75         if (s[i] >= 90)
 76             r++;
 77         if (s[i] < 60)
 78             p++;
 79     }
 80     x = r; y = p;
 81 
 82 }
 83 int main()
 84 {
 85     int s[N];
 86     cout << "输入学生人数：";
 87     int n,i, x, y;
 88     cin >> n;
 89     cout << "输入学生成绩（满分100）：";
 90     for (i = 0; i < n; i++)
 91     {
 92         cin >> s[i];
 93     }
 94     fun2(s, n, x,y);
 95     cout << "优秀人数为" << x << endl;
 96     cout << "不及格人数为" << y << endl;
 97     cout << "从高到低排序成绩：" << endl;
 98     for (i = 0; i < n; i++)
 99     {
100         cout << s[i]<<' ';
101     }
102     system("pause");
103     return 0;
104 }
View Code

 

cm's best ver:

  1 #include "iostream"
  2 
  3 #include <string.h>
  4 
  5 using namespace std;
  6 
  7 
  8 
  9 int fun1(int s[],int n,int *x);
 10 
 11 void fun2(int s[],int n,int &x,int &y);
 12 
 13 int main()
 14 
 15 {
 16 
 17     const int N = 100;
 18 
 19     int s[N] = {0};
 20 
 21     int n;
 22 
 23     cin>>n;
 24 
 25     for(int i=0; i<n; ++i)
 26 
 27         cin>>s[i];
 28 
 29 
 30 
 31     int x=0,y=0;
 32 
 33     cout<<"优秀的人数："<<fun1(s,n,&x)<<endl;
 34 
 35     cout<<"不及格的人数："<<x<<endl;
 36 
 37 
 38 
 39     x=0,y=0;
 40 
 41     fun2(s,n,x,y);
 42 
 43     cout<<"优秀的人数："<<y<<endl<<"不及格的人数："<<x<<endl;
 44 
 45 
 46 
 47     return 0;
 48 
 49 }
 50 
 51 int fun1(int s[],int n,int *x){
 52 
 53     int excellent=0;
 54 
 55     * x = 0;//
 56 
 57     for(int i=0; i<n-1; ++i){/*第0个元素有序，从第1个元素向右无序*/
 58 
 59         int max = s[i];
 60 
 61         int index = i;
 62 
 63         for(int j = i+1; j<n; ++j){/*从i+1逐个比较*/
 64 
 65             if(max<s[j]){ /*是否比后面的小*/
 66 
 67                 max = s[j];
 68 
 69                 index = j;
 70 
 71             }
 72 
 73         }
 74 
 75         if(index != i){/*找到了最大值才交换*/
 76 
 77             s[index] = s[i];
 78 
 79             s[i] = max;
 80 
 81         }
 82 
 83     }
 84 
 85     for(int k=0; k<n; ++k){
 86 
 87         if(s[k]>=90) excellent++;
 88 
 89         if(s[k]<60) (*x)++;
 90 
 91     }
 92 
 93     return excellent;
 94 
 95 }
 96 
 97 void fun2(int s[],int n,int &x,int &y){
 98 
 99         x=y=0;  //
100 
101     for(int i=0; i<n-1; ++i){/*第0个元素有序，从第1个元素向右无序*/
102 
103         int max = s[i];
104 
105         int index = i;
106 
107         for(int j = i+1; j<n; ++j){/*从i+1逐个比较*/
108 
109             if(max<s[j]){ /*是否比后面的小*/
110 
111                 max = s[j];
112 
113                 index = j;
114 
115             }
116 
117         }
118 
119         if(index != i){/*找到了最大值才交换*/
120 
121             s[index] = s[i];
122 
123             s[i] = max;
124 
125         }
126 
127     }
128 
129     for(int k=0; k<n; ++k){
130 
131         if(s[k]>=90) y++;
132 
133         if(s[k]<60) x++;
134 
135     }
136 
137 }
View Code

 or:

 1 #include "iostream"
 2 using namespace std;
 3 
 4 int fun1(int s[], int n, int* x)
 5 {
 6     int i, j, k, good = 0;
 7     *x = 0;
 8     for (i = 0; i <= n - 1; i++)
 9     {
10         for (j = n - 1; j > i; j--)
11             if (s[j] > s[j - 1])
12             {
13                 k = s[j];
14                 s[j] = s[j - 1];
15                 s[j - 1] = k;
16             }
17     }
18     for (i = 0; i <= n - 1; i++)
19     {
20         if (s[i] >= 90)
21             good += 1;
22         else if (s[i] <= 60)
23             *x += 1;
24     }
25     return good;
26 }
27 
28 void fun2(int s[], int n, int& x, int& y)
29 {
30     int i, j, k;
31     x = y = 0;
32     for (i = 0; i <= n - 1; i++)
33     {
34         for (j = n - 1; j > i; j--)
35             if (s[j] > s[j - 1])
36             {
37                 k = s[j];
38                 s[j] = s[j - 1];
39                 s[j - 1] = k;
40             }
41     }
42     for (i = 0; i <= n - 1; i++)
43     {
44         if (s[i] >= 90)
45             x += 1;
46         else if (s[i] <= 60)
47             y += 1;
48     }
49 }
50 
51 int main()
52 {
53     //数据输入
54     int n, s[100], good, bad;
55     cout << "请输入学生人数:" << endl;
56     cin >> n;
57     cout << "请输入学生成绩:" << endl;
58     for (int i = 0; i <= n - 1; i++)
59         cin >> s[i];
60 
61     //调用fun1
62     good = fun1(s, n, &bad);
63     cout << "排序后的成绩为:" << endl;
64     for (int i = 0; i <= n - 1; i++)
65         cout << s[i] << " ";
66     cout << endl << "优秀人数为:" << good << endl;
67     cout << "不及格人数为:" << bad << endl;
68 
69     //调用fun2
70     fun2(s, n, good, bad);
71     cout << "排序后的成绩为:" << endl;
72     for (int i = 0; i <= n - 1; i++)
73         cout << s[i] << " ";
74     cout << endl << "优秀人数为:" << good << endl;
75     cout << "不及格人数为:" << bad << endl;
76 
77     system("pause");
78     return 0;
79 }
View Code

 

4. 

编一函数，功能为统计字符串中各个字母（不区分大、小写）出现的频率，同时找出频率出现最高的字母及次数，假设出现次数最多的字母只有一个。函数形式为：

void freq(char s[],int p[],char &chmax,int &max)

程序运行结果如下：

yzy.ver:

 1 #include "iostream"
 2 #define N 256
 3 using namespace std;
 4 
 5 void freq(char s[], int p[], char& chmax, int& max)
 6 {
 7     char *a = s;
 8     while(*a)
 9     {
10         if (*a > 'a' && *a < 'z' || *a>'A' && *a < 'Z')
11         {
12             if (*a < 'a')
13                 p[*a + 32]++;
14             else
15                 p[*a]++;
16             if (p[*a] > max)
17             {
18                 max = p[*a];
19                 chmax = *a;
20             }
21         }
22         a++;
23     }
24 }
25 
26 
27 
28 int main()
29 {
30     char s[N] = "" ;
31     char chmax ;
32     int p[N]={}, max=0;
33     gets_s(s);
34     freq(s, p, chmax, max);
35     for (int i = 0; i < N; i++)
36     {
37         if (p[i])
38             cout << char(i) << "----" << p[i] << endl;
39     }
40     cout << "出现频率最高的字母：" << chmax << "----" << max << endl;
41     system("pause");
42     return 0;
43 }
View Code

 

cm's best ver:

 1 #include "iostream"
 2 
 3 #include <string.h>
 4 
 5 using namespace std;
 6 
 7 
 8 
 9 void freq(char s[],int p[],char &chmax,int &max);
10 
11 int main()
12 
13 {
14 
15     char chmax;
16 
17     int max = 0;
18 
19     const int N = 256;
20 
21     int p[N] = {0};
22 
23     char s[N] = "";
24 
25     gets(s);
26 
27     freq(s,p,chmax,max);
28 
29     for(int i=0; i<N; ++i){
30 
31         if(p[i])
32 
33             printf("%c-----%d\n",i,p[i]);
34 
35     }
36 
37     printf("出现频率最高的字母:%c-----%d\n",chmax,max);
38 
39     return 0;
40 
41 }
42 
43 void freq(char s[],int p[],char &chmax,int &max){
44 
45     char *q = s;
46 
47     while(*q)
48 
49     {
50 
51         if(*q>='a'&&*q<='z'||*q>='A'&&*q<='Z')
52 
53         {
54 
55             if(*q<'a')
56 
57                 p[*q+32]++;
58 
59             else
60 
61                 p[*q]++;
62 
63             if(p[*q]>max){
64 
65                 max = p[*q];
66 
67                 chmax = *q;
68 
69             }
70 
71         }
72 
73         q++;
74 
75     }
76 
77 }
View Code

 

posted @ 2023-11-23 10:29  YANTARES  阅读(99)  评论(0)    收藏  举报