C程序设计(第5版)谭浩强习题解答
第7章 用函数实现模块化程序设计
1.写两个函数,分别求两个整数的最大公约数和最小公倍数,用主函数调用这两个函数,并输出结果。两个整数由键盘输人。
//7.1.1
#include
int main()
{
int hcf(int, int);
int lcd(int, int, int);
int u, v, h, l;
scanf("%d,%d", &u, &v);
h = hcf(u, v);
printf("H.C.F=%d\n", h);
l = lcd(u, v, h);
printf("L.C.D=%d\n", l);
return 0;
}
int hcf(int u, int v)
{
int t, r;
if (v > u)
{
t = u; u = v; v = t;
}
while ((r = u % v) != 0)
{
u = v;
v = r;
}
return(v);
}
int lcd(int u, int v, int h)
{
return(u*v / h);
}
//7.1.2
#include
int Hcf, Lcd;
int main()
{
void hcf(int, int);
void lcd(int, int);
int u, v;
scanf("%d,%d", &u, &v);
hcf(u, v);
lcd(u, v);
printf("H.C.F=%d\n", Hcf);
printf("L.C.D=%d\n", Lcd);
return 0;
}
void hcf(int u, int v)
{
int t, r;
if (v > u)
{
t = u; u = v; v = t;
}
while ((r = u % v) != 0)
{
u = v;
v = r;
}
Hcf = v;
}
void lcd(int u, int v)
{
Lcd = u * v / Hcf;
}
2.求方程 ax2+bx+c=0的根,用3个函数分别求当:b2−4ac大于0、等于0和小于0时的根并输出结果。从主函数输入a,b,c的值。
#include
#include
float x1, x2, disc, p, q;
int main()
{
void greater_than_zero(float, float);
void equal_to_zero(float, float);
void smaller_than_zero(float, float);
float a, b, c;
printf("input a,b,c:");
scanf("%f,%f,%f", &a, &b, &c);
printf("equation: %5.2f*x*x+%5.2f*x+%5.2f=0\n", a, b, c);
disc = b * b - 4 * a*c;
printf("root:\n");
if (disc > 0)
{
greater_than_zero(a, b);
printf("x1=%f\t\tx2=%f\n", x1, x2);
}
else if (disc == 0)
{
equal_to_zero(a, b);
printf("x1=%f\t\tx2=%f\n", x1, x2);
}
else
{
smaller_than_zero(a, b);
printf("x1=%f+%fi\tx2=%f-%fi\n", p, q, p, q);
}
return 0;
}
void greater_than_zero(float a, float b)
{
x1 = (-b + sqrt(disc)) / (2 * a);
x2 = (-b - sqrt(disc)) / (2 * a);
}
void equal_to_zero(float a, float b)
{
x1 = x2 = (-b) / (2 * a);
}
void smaller_than_zero(float a, float b)
{
p = -b / (2 * a);
q = sqrt(-disc) / (2 * a);
}
3.写一个判素数的函数,在主函数输人一个整数,输出是否为素数的信息。
#include
int main()
{
int prime(int);
int n;
printf("input an integer:");
scanf("%d", &n);
if (prime(n))
printf("%d is a prime.\n", n);
else
printf("%d is not a prime.\n", n);
return 0;
}
int prime(int n)
{
int flag = 1, i;
for (i = 2; i < n / 2 && flag == 1; i++)
if (n%i == 0)
flag = 0;
return(flag);
}
4.写一个函数,使给定的一个3X3的二维整型数组转置,即行列互换。
#include
#define N 3
int array[N][N];
int main()
{
void convert(int array[][3]);
int i, j;
printf("input array:\n");
for (i = 0; i < N; i++)
for (j = 0; j < N; j++)
scanf("%d", &array[i][j]);
printf("\noriginal array :\n");
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
printf("%5d", array[i][j]);
printf("\n");
}
convert(array);
printf("convert array:\n");
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
printf("%5d", array[i][j]);
printf("\n");
}
return 0;
}
void convert(int array[][3])
{
int i, j, t;
for (i = 0; i < N; i++)
for (j = i + 1; j < N; j++)
{
t = array[i][j];
array[i][j] = array[j][i];
array[j][i] = t;
}
}
5.写一个函数,使输人的一个字符串按反序存放,在主函数中输入和输出字符串。
#include
#include
int main()
{
void inverse(char str[]);
char str[100];
printf("input string:");
scanf("%s", str);
inverse(str);
printf("inverse string:%s\n", str);
return 0;
}
void inverse(char str[])
{
char t;
int i, j;
for (i = 0, j = strlen(str); i < (strlen(str) / 2); i++, j--)
{
t = str[i];
str[i] = str[j - 1];
str[j - 1] = t;
}
}
6.写一个函数,将两个字符串连接。
#include
int main()
{
void concatenate(char string1[], char string2[], char string[]);
char s1[100], s2[100], s[100];
printf("input string1:");
scanf("%s", s1);
printf("input string2:");
scanf("%s", s2);
concatenate(s1, s2, s);
printf("\nThe new string is %s\n", s);
return 0;
}
void concatenate(char string1[], char string2[], char string[])
{
int i, j;
for (i = 0; string1[i] != '\0'; i++)
string[i] = string1[i];
for (j = 0; string2[j] != '\0'; j++)
string[i + j] = string2[j];
string[i + j] = '\0';
}
7.写一个函数,将一个字符串中的元音字母复制到另一字符串,然后输出。
#include
int main()
{
void cpy(char[], char[]);
char str[80], c[80];
printf("input string:");
gets(str);
cpy(str, c);
printf("The vowel letters are:%s\n", c);
return 0;
}
void cpy(char s[], char c[])
{
int i, j;
for (i = 0, j = 0; s[i] != '\0'; i++)
if (s[i] == 'a' || s[i] == 'A' || s[i] == 'e' || s[i] == 'E' || s[i] == 'i' ||
s[i] == 'I' || s[i] == 'o' || s[i] == 'O' || s[i] == 'u' || s[i] == 'U')
{
c[j] = s[i];
j++;
}
c[j] = '\0';
}
8.写一个函数,输人一个4位数字，要求输出这4个数字字符,但每两个数字间空一个空格。如输人1990,应输出“1 9 9 0”。
#include
#include
int main()
{
char str[80];
void insert(char[]);
printf("input four digits:");
scanf("%s", str);
insert(str);
return 0;
}
void insert(char str[])
{
int i;
for (i = strlen(str); i > 0; i--)
{
str[2 * i] = str[i];
str[2 * i - 1] = ' ';
}
printf("output:\n%s\n", str);
}
9.编写一个函数,由实参传来一个字符串,统计此字符串中字母、数字、空格和其他字符的个数,在主函数中输人字符串以及输出上述的结果。
#include
int letter, digit, space, others;
int main()
{
void count(char[]);
char text[80];
printf("input string:\n");
gets(text);
printf("string:");
puts(text);
letter = 0;
digit = 0;
space = 0;
others = 0;
count(text);
printf("\nletter:%d\ndigit:%d\nspace:%d\nothers:%d\n", letter, digit, space, others);
return 0;
}
void count(char str[])
{
int i;
for (i = 0; str[i] != '\0'; i++)
if ((str[i] >= 'a'&& str[i] <= 'z') || (str[i] >= 'A' && str[i] <= 'Z'))
letter++;
else if (str[i] >= '0' && str[i] <= '9')
digit++;
else if (str[i] == 32)
space++;
else
others++;
}
10.写一个函数,输人一行字符,将此字符串中最长的单词输出。
#include
#include
int main()
{
int alphabetic(char);
int longest(char[]);
int i;
char line[100];
printf("input one line:\n");
gets(line);
printf("The longest word is :");
for (i = longest(line); alphabetic(line[i]); i++)
printf("%c", line[i]);
printf("\n");
return 0;
}
int alphabetic(char c)
{
if ((c >= 'a' && c <= 'z') || (c >= 'A'&&c <= 'z'))
return(1);
else
return(0);
}
int longest(char string[])
{
int len = 0, i, length = 0, flag = 1, place = 0, point;
for (i = 0; i <= strlen(string); i++)
if (alphabetic(string[i]))
if (flag)
{
point = i;
flag = 0;
}
else
len++;
else
{
flag = 1;
if (len >= length)
{
length = len;
place = point;
len = 0;
}
}
return(place);
}
11.写一个函数,用“起泡法”对输人的10个字符按由小到大顺序排列。
#include
#include
#define N 10
char str[N];
int main()
{
void sort(char[]);
int i, flag;
for (flag = 1; flag == 1;)
{
printf("input string:\n");
scanf("%s", &str);
if (strlen(str) > N)
printf("string too long,input again!");
else
flag = 0;
}
sort(str);
printf("string sorted:\n");
for (i = 0; i < N; i++)
printf("%c", str[i]);
printf("\n");
return 0;
}
void sort(char str[])
{
int i, j;
char t;
for (j = 1; j < N; j++)
for (i = 0; (i < N - j) && (str[i] != '\0'); i++)
if (str[i] > str[i + 1])
{
t = str[i];
str[i] = str[i + 1];
str[i + 1] = t;
}
}
12.用牛顿迭代法求根。方程为ax3+bx2+cx+d=0,系数a,b,c,d的值依次为1,2,3,4,由主函数输人。求x在1附近的一个实根。求出根后由主函数输出。
#include
#include
int main()
{
float solut(float a, float b, float c, float d);
float a, b, c, d;
printf("input a,b,c,d:");
scanf("%f,%f,%f,%f", &a, &b, &c, &d);
printf("x=%10.7f\n", solut(a, b, c, d));
return 0;
}
float solut(float a, float b, float c, float d)
{
float x = 1, x0, f, f1;
do
{
x0 = x;
f = ((a*x0 + b)*x0 + c)*x0 + d;
f1 = (3 * a*x0 + 2 * b)*x0 + c;
x = x0 - f / f1;
} while (fabs(x - x0) >= 1e-3);
return(x);
}
13.用递归方法求n阶勒让德多项式的值,递归公式为
#include
#define N 10
#define M 5
float score[N][M];
float a_stu[N], a_cour[M];
int r, c;
int main()
{
int i, j;
float h;
float s_var(void);
float highest();
void input_stu(void);
void aver_stu(void);
void aver_cour(void);
input_stu();
aver_stu();
aver_cour();
printf("\n NO. cour1 cour2 cour3 cour4 cour5 aver\n");
for (i = 0; i < N; i++)
{
printf("\n NO %2d ", i + 1);
for (j = 0; j < M; j++)
printf("%8.2f", score[i][j]);
printf("%8.2f\n", a_stu[i]);
}
printf("\naverage:");
for (j = 0; j < M; j++)
printf("%8.2f", a_cour[j]);
printf("\n");
h = highest();
printf("highest:%7.2f NO. %2d course %2d\n", h, r, c);
printf("variance %8.2f\n", s_var());
return 0;
}
void input_stu(void)
{
int i, j;
for (i = 0; i < N; i++)
{
printf("\ninput score of student%2d:\n", i + 1);
for (j = 0; j < M; j++)
scanf("%f", &score[i][j]);
}
}
void aver_stu(void)
{
int i, j;
float s;
for (i = 0; i < N; i++)
{
for (j = 0, s = 0; j < M; j++)
s += score[i][j];
a_stu[i] = s / 5.0;
}
}
void aver_cour(void)
{
int i, j;
float s;
for (j = 0; j < M; j++)
{
s = 0;
for (i = 0; i < N; i++)
s += score[i][j];
a_cour[j] = s / (float)N;
}
}
float highest()
{
float high;
int i, j;
high = score[0][0];
for (i = 0; i < N; i++)
for (j = 0; j < M; j++)
if (score[i][j] > high)
{
high = score[i][j];
r = i + 1;
c = j + 1;
}
return(high);
}
float s_var(void)
{
int i;
float sumx, sumxn;
sumx = 0.0;
sumxn = 0.0;
for (i = 0; i < N; i++)
{
sumx += a_stu[i] * a_stu[i];
sumxn += a_stu[i];
}
return(sumx / N - (sumxn / N)*(sumxn / N));
}
14.输人10个学生5门课的成绩,分别用函数实现下列功能:
①计算每个学生的平均分;
②计算每门课的平均分;
③找出所有50个分数中最高的分数所对应的学生和课程;
④计算平均分方差:
其中,x;为某一学生的平均分。
#include
#define N 10
#define M 5
float score[N][M];
float a_stu[N], a_cour[M];
int r, c;
int main()
{
int i, j;
float h;
float s_var(void);
float highest();
void input_stu(void);
void aver_stu(void);
void aver_cour(void);
input_stu();
aver_stu();
aver_cour();
printf("\n NO. cour1 cour2 cour3 cour4 cour5 aver\n");
for (i = 0; i < N; i++)
{
printf("\n NO %2d ", i + 1);
for (j = 0; j < M; j++)
printf("%8.2f", score[i][j]);
printf("%8.2f\n", a_stu[i]);
}
printf("\naverage:");
for (j = 0; j < M; j++)
printf("%8.2f", a_cour[j]);
printf("\n");
h = highest();
printf("highest:%7.2f NO. %2d course %2d\n", h, r, c);
printf("variance %8.2f\n", s_var());
return 0;
}
void input_stu(void)
{
int i, j;
for (i = 0; i < N; i++)
{
printf("\ninput score of student%2d:\n", i + 1);
for (j = 0; j < M; j++)
scanf("%f", &score[i][j]);
}
}
void aver_stu(void)
{
int i, j;
float s;
for (i = 0; i < N; i++)
{
for (j = 0, s = 0; j < M; j++)
s += score[i][j];
a_stu[i] = s / 5.0;
}
}
void aver_cour(void)
{
int i, j;
float s;
for (j = 0; j < M; j++)
{
s = 0;
for (i = 0; i < N; i++)
s += score[i][j];
a_cour[j] = s / (float)N;
}
}
float highest()
{
float high;
int i, j;
high = score[0][0];
for (i = 0; i < N; i++)
for (j = 0; j < M; j++)
if (score[i][j] > high)
{
high = score[i][j];
r = i + 1;
c = j + 1;
}
return(high);
}
float s_var(void)
{
int i;
float sumx, sumxn;
sumx = 0.0;
sumxn = 0.0;
for (i = 0; i < N; i++)
{
sumx += a_stu[i] * a_stu[i];
sumxn += a_stu[i];
}
return(sumx / N - (sumxn / N)*(sumxn / N));
}
15.写几个函数:
①输人10个职工的姓名和职工号;
②按职工号由小到大顺序排序,姓名顺序也随之调整;
③要求输人一个职工号,用折半查找法找出该职工的姓名,从主函数输人要查找的职工号,输出该职工姓名。
#include
#include
#define N 10
int main()
{
void input(int[], char name[][8]);
void sort(int[], char name[][8]);
void search(int, int[], char name[][8]);
int num[N], number, flag = 1, c;
char name[N][8];
input(num, name);
sort(num, name);
while (flag == 1)
{
printf("\ninput number to look for:");
scanf("%d", &number);
search(number, num, name);
printf("continue ot not(Y/N)?");
getchar();
c = getchar();
if (c == 'N' || c == 'n')
flag = 0;
}
return 0;
}
void input(int num[], char name[N][8])
{
int i;
for (i = 0; i < N; i++)
{
printf("input NO.: ");
scanf("%d", &num[i]);
printf("input name: ");
getchar();
gets(name[i]);
}
}
void sort(int num[], char name[N][8])
{
int i, j, min, templ;
char temp2[8];
for (i = 0; i < N - 1; i++)
{
min = i;
for (j = i; j < N; j++)
if (num[min] > num[j]) min = j;
templ = num[i];
strcpy(temp2, name[i]);
num[i] = num[min];
strcpy(name[i], name[min]);
num[min] = templ;
strcpy(name[min], temp2);
}
printf("\n result:\n");
for (i = 0; i < N; i++)
printf("\n %5d%10s", num[i], name[i]);
}
void search(int n, int num[], char name[N][8])
{
int top, bott, mid, loca, sign;
top = 0;
bott = N - 1;
loca = 0;
sign = 1;
if ((n < num[0]) || (n > num[N - 1]))
loca = -1;
while ((sign == 1) && (top <= bott))
{
mid = (bott + top) / 2;
if (n == num[mid])
{
loca = mid;
printf("NO. %d , his name is %s.\n", n, name[loca]);
sign = -1;
}
else if (n < num[mid])
bott = mid - 1;
else
top = mid + 1;
}
if (sign == 1 || loca == -1)
printf("%d not been found.\n", n);
}
16.写一个函数,输人一个十六进制数,输出相应的十进制数。
#include
#define MAX 1000
int main()
{
int htoi(char s[]);
int c, i, flag, flag1;
char t[MAX];
i = 0;
flag = 0;
flag1 = 1;
printf("input a HEX number:");
while ((c = getchar()) != '\0' && i < MAX&& flag1)
{
if (c >= '0' && c <= '9' || c >= 'a' && c <= 'f' || c >= 'A' && c <= 'F')
{
flag = 1;
t[i++] = c;
}
else if (flag)
{
t[i] = '\0';
printf("decimal number %d\n", htoi(t));
printf("continue or not?");
c = getchar();
if (c == 'N' || c == 'n')
flag1 = 0;
else
{
flag = 0;
i = 0;
printf("\ninput a HEX number:");
}
}
}
return 0;
}
int htoi(char s[])
{
int i, n;
n = 0;
for (i = 0; s[i] != '\0'; i++)
{
if (s[i] >= '0'&& s[i] <= '9')
n = n * 16 + s[i] - '0';
if (s[i] >= 'a' && s[i] <= 'f')
n = n * 16 + s[i] - 'a' + 10;
if (s[i] >= 'A' && s[i] <= 'F')
n = n * 16 + s[i] - 'A' + 10;
}
return(n);
}
17.用递归法将一个整数n转换成字符串。例如，输人483,应输出字符串”483”。n的位数不确定,可以是任意位数的整数。
#include
int main()
{
void convert(int n);
int number;
printf("input an integer: ");
scanf("%d", &number);
printf("output: ");
if (number < 0)
{
putchar('-'); putchar(' '); //先输出一个‘-’号和空格
number = -number;
}
convert(number);
printf("\n");
return 0;
}
void convert(int n)
{
int i;
if ((i = n / 10) != 0)
convert(i);
putchar(n % 10 + '0');
putchar(32);
}
18.给出年、月、日,计算该日是该年的第几天。
#include
int main()
{
int sum_day(int month, int day);
int leap(int year);
int year, month, day, days;
printf("input date(year,month,day):");
scanf("%d,%d,%d", &year, &month, &day);
printf("%d/%d/%d ", year, month, day);
days = sum_day(month, day); //调用函数sum_day
if (leap(year) && month >= 3) //调用函数leap
days = days + 1;
printf("is the %dth day in this year.\n", days);
return 0;
}
int sum_day(int month, int day) //函数sum_day:计算日期
{
int day_tab[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
int i;
for (i = 1; i < month; i++)
day += day_tab[i]; //累加所在月之前天数
return(day);
} //函数leap:判断是否为闰年
int leap(int year)
{
int leap;
leap = year % 4 == 0 && year % 100 != 0 || year % 400 == 0;
return(leap);
}
