山科C语言考试资料参考

1.数列反向 1 2 3编程3 2 1

/* Reverses a series of numbers */

#include

#define N 10

int main(void)

{

int a[N], i;

printf("Enter %d numbers: ", N);

for (i = 0; i < N; i++)

scanf("%d", &a[i]);

printf("In reverse order:");

for (i = N - 1; i >= 0; i--)

printf(" %d", a[i]);

printf("\n");

return 0;

}

1.检查数中重复出现的数字

/* Checks numbers for repeated digits */ #include /* C99 only */ #include

int main(void)

{

bool digit_seen[10] = {false};

int digit;

long n;

printf("Enter a number: ");

scanf("%ld", &n);

while (n > 0) {

digit = n % 10;

if (digit_seen[digit])

break;

digit_seen[digit] = true;

n /= 10;

}

if (n > 0)

printf("Repeated digit\n");

else

printf("No repeated digit\n");

return 0;

}

2.计算利息

/* Prints a table of compound interest */

#include

#define NUM_RATES ((int) (sizeof(value) / sizeof(value[0]))) #define INITIAL_BALANCE 100.00

int main(void)

{

int i, low_rate, num_years, year;

double value[5];

printf("Enter interest rate: ");

scanf("%d", &low_rate);

printf("Enter number of years: ");

scanf("%d", &num_years);

printf("\nYears");

for (i = 0; i < NUM_RATES; i++) {

printf("%6d%%", low_rate + i);

value[i] = INITIAL_BALANCE;

}

printf("\n");

for (year = 1; year <= num_years; year++) {

printf("%3d ", year);

for (i = 0; i < NUM_RATES; i++) {

value[i] += (low_rate + i) / 100.0 * value[i];

printf("%7.2f", value[i]);

}

printf("\n");

}

return 0;

}

4.发牌

/* Deals a random hand of cards */

#include /* C99 only */

#include

#include

#include

#define NUM_SUITS 4

#define NUM_RANKS 13

int main(void)

{

bool in_hand[NUM_SUITS][NUM_RANKS] = {false};

int num_cards, rank, suit;

const char rank_code[] = {'2','3','4','5','6','7','8', '9','t','j','q','k','a'};

const char suit_code[] = {'c','d','h','s'};

srand((unsigned) time(NULL));

printf("Enter number of cards in hand: ");

scanf("%d", &num_cards);

printf("Your hand:");

while (num_cards > 0) {

suit = rand() % NUM_SUITS; /* picks a random suit */ rank = rand() % NUM_RANKS; /* picks a random rank */ if (!in_hand[suit][rank]) {

in_hand[suit][rank] = true;

num_cards--;

printf(" %c%c", rank_code[rank], suit_code[suit]);

}

}

printf("\n");

return 0;

}

5冒泡排序

void main() {

int a[10],k,j,t;

printf("输入10个整数：\n");

for( k = 0; k < 10; k ++ )

scanf("%d",&a[ k ]); //依次输入10个整数

for( j = 0; j < 9; j ++ ) //进行9轮排序即n－1次

{

for( k = 0; k < 9-j; k ++) //每轮进行n－1－j 次比较

if( a[ k ] > a[ k + 1 ] )

{t=a[k]; a[k]=a[k+1]; a[k+1]=t;} //小的沉底，大的上浮 }

printf("排序结果:");

for( k = 0; k < 10; k ++ ) //依次输出排序结果

printf("%d\t",a[ k ]);

printf("\n");

}

6.判定素数

/* Tests whether a number is prime */

#include /* C99 only */

#include

bool is_prime(int n)

{

int divisor;

if (n <= 1)

return false;

for (divisor = 2; divisor * divisor <= n; divisor++)

if (n % divisor == 0)

return false;

return true;

}

int main(void)

{

int n;

printf("Enter a number: ");

scanf("%d", &n);

if (is_prime(n))

printf("Prime\n");

else

printf("Not prime\n");

return 0;

}

7.汉诺塔问题

void move(char getone, char putone) /*打印移动信息*/ { printf("%c--->%c\n",getone,putone); }

void hanoi(int n,char one,char two,char three) /*递归函数*/ { if(n==1) move(one,three);

else

{ hanoi(n-1,one,three,two);

move(one,three);

hanoi(n-1,two,one,three); }}

main()

{ int m;

printf("Input the number of disks:");

scanf("%d",&m);

printf("The steps to moving %3d disks:\n",m);

hanoi(m,'A','B','C'); }

8.猜数游戏自动产生一个1-100之间的随机数，提示用户尝试用尽可能少的次数猜出这个数/* Asks user to guess a hidden number */

#include

#include

#include

#define MAX_NUMBER 100

/* prototypes */

void initialize_number_generator(void);

int new_secret_number(void);

void read_guesses(int secret_number);

int main(void)

{

char command;

int secret_number;

printf("Guess the secret number between 1 and %d.\n\n",MAX_NUMBER); initialize_number_generator();

do {

secret_number = new_secret_number();

printf("A new number has been chosen.\n");

read_guesses(secret_number);

printf("Play again? (Y/N) ");

scanf(" %c", &command);

printf("\n");

} while (command == 'y' || command == 'Y');

return 0;

}

9.找出数组中最大和最小元素

/* Finds the largest and smallest elements in an array */ #include

#define N 10

void max_min(int a[], int n, int *max, int *min);

int main(void)

{

int b[N], i, big, small;

printf("Enter %d numbers: ", N);

for (i = 0; i < N; i++)

scanf("%d", &b[i]);

max_min(b, N, &big, &small);

printf("Largest: %d\n", big);

printf("Smallest: %d\n", small);

return 0;

}

void max_min(int a[], int n, int *max, int *min)

{

int i;

*max = *min = a[0];

for (i = 1; i < n; i++) {

if (a[i] > *max)

*max = a[i];

else if (a[i] < *min)

*min = a[i];

}

}

第八章的程序reverse.c实现了读入10个数，然后对其进行逆序输出第1题的升级版

/* Reverses a series of numbers (pointer version) */

#include

#define N 10

int main(void)

{

int a[N], *p;

printf("Enter %d numbers: ", N);

for (p = a; p < a + N; p++)

scanf("%d", p);

printf("In reverse order:");

for (p = a + N - 1; p >= a; p--)

printf(" %d", *p);

printf("\n");

return 0;

}

第16组

例：对候选人得票的统计程序。设有3个候选人，每次输入一个得票的候选人的名字，要求最后输出各人得票结果。

#include

#include

struct person

{

char name[20];in count;

};

leader[3]={“Li”,0,”Zhang”,0,”Fun”,0}

void main()

{ int i,j; char leader_name[20];

for(i=1;i<=10;i++)

{

scanf(“%s”,leader_name);

for(j=0;j<3;j++)

if(strcmp(leader_name,leader[j].name)==0)

leader[j].count++;

}

printf(“\n”);

for(i=0;i<3;i++) printf(“%5s:%d\n”,leader[i].name,leader[i].count);} 例：有一个结构体变量stu，内含学生学号、姓名和3门课程的成绩。要求在main函数中赋予值，在另一函数print中将它们输出。今用结构体变量作函数参数。

#include

struct student

{

int num;

char name[20];

float score[3];

};

void main()

{

void print(struct student);

struct student stu;

stu.num=12345;strcpy(https://www.360docs.net/doc/8d4035601.html,, ″LiLin″;stu.score[0]=67.5;stu.score[1]=89;stu.score[2]

=78.6);

print(stu);

}

void print(struct student stu)

{ printf(FORMAT，stu.num，https://www.360docs.net/doc/8d4035601.html,， stu.score[0]， stu.score[1]，stu.score[2]）； printf（″\n″）；}

例：将上题改用指向结构体变量的指针作实参。

#include

struct student

{

int num;

char name[20];

float score[3];

};stu={12345, ″LiLi″,67.5,89,78.6};

void main()

{void print(struct student *);/*形参类型修改成指向结构体的指针变量*/

print(&stu);} /*实参改为stu的起始地址*/

void print(struct student *p) /*形参类型修改了*/

{ printf(FORMAT，p->num，p->name，

p->score［0］，p->score［1］，p->score［2］）； /*用指针变量调用各成员的值*/

printf（″＼ｎ″）；}

例:设有若干个人员的数据，其中有学生和教师。学生的数据中包括：姓名、号码、性别、职业、班级。教师的数据包括：姓名、号码、性别、职业、职务。可以看出，学生和教师所包含的数据是不同的。现要求把它们放在同一表格中。

#include

struct

{

int num;

char name[10];

char sex;

char job;

union

{ int banji;

char position[10];

}category;

}person[2];/*先设人数为2*/

void main()

{int i;

for(i=0;i<2;i++)

{scanf("%d %s %c %c", &person[i].num, &person[i].name,

&person[i].sex, &person[i].job);

if(person[i].job == 'S')

scanf("%d", &person[i].category.banji);

else if(person[i].job == 'T')

scanf("%s", person[i].category.position);

else printf(“Input error!”);} printf("\n");

printf("No. name sex job class/position\n");

for(i=0;i<2;i++)

{if (person[i].job == 'S')

printf(“%-6d%-10s%-3c%-3c%-6d\n”,person[i].num, person[i].name, person[i].sex, person[i].job, person[i].category.banji);

else printf(“%-6d%-10s%-3c%-3c%-6s\n”,person[i].num,

person[i].name,person[i].sex, person[i].job,

person[i].category.position);}}