1.
#include <stdio.h>
double min (double x, double y);
int main (void)
{
double x, y;
printf ("Enter two numbers of double:\n");
scanf ("%lf %lf", &x, &y);
printf ("%lf", min (x, y));
return 0;
}
double min (double a, double b)
{
double c = a;
if (a > b)
c = b;
return c;
}
2.
#include <stdio.h>
void chline (char ch, int i, int j);
int main (void)
{
int i, j;
char ch;
printf ("Enter a word:\n");
scanf ("%c", &ch);
printf ("Enter the number of column and line:\n");
scanf ("%d %d", &i, &j);
chline (ch, i, j);
return 0;
}
void chline (char ch, int i, int j)
{
int a, b;
for (a = 0; a < j; a++)
{
for (b = 0; b < i; b++)
printf ("%c", ch);
printf ("\n");
}
}
3.
第三题与第二题题目描述一致
#include <stdio.h>
void chline (char ch, int i, int j);
int main (void)
{
int i, j;
char ch;
printf ("Enter a word:\n");
scanf ("%c", &ch);
printf ("Enter the number of column and line:\n");
scanf ("%d %d", &i, &j);
chline (ch, i, j);
return 0;
}
void chline (char ch, int i, int j)
{
int a, b;
for (a = 0; a < j; a++)
{
for (b = 0; b < i; b++)
printf ("%c", ch);
printf ("\n");
}
}
4.
#include <stdio.h>
double aver (double a, double b);
int main (void)
{
double i, j;
printf ("Enter two numbers of double:\n");
scanf ("%lf %lf", &i, &j);
printf ("%lf", aver (i, j));
return 0;
}
double aver (double i, double j)
{
double a = (1 / j + 1 / i) / 2;
return 1 / a;
}
5.
#include <stdio.h>
void larger_of (double, double);
int main (void)
{
double a, b;
printf ("Enter two numbers of double:\n");
scanf ("%lf %lf", &a, &b);
larger_of (a, b);
return 0;
}
larger_of (double a, double b)
{
if (a > b)
printf ("a = b = %lf", a);
else
printf ("a = b = %lf", b);
}
6.
#include <stdio.h>
void com (double*, double*, double*);
int main (void)
{
double a, b, c;
printf ("Enter three numbers of double:\n");
scanf ("%lf %lf %lf", &a, &b, &c);
com (&a, &b, &c);
printf ("a = %lf\nb = %lf\nc = %lf", a, b, c);
return 0;
}
void com (double* a, double* b, double* c)
{
double t;
if (*a > *b)
{
t = *a;
*a = *b;
*b = t;
}
if (*a > *c)
{
t = *a;
*a = *c;
*c = t;
}
if (*b > *c)
{
t = *b;
*b = *c;
*c = t;
}
}
7.
#include <stdio.h>
int loca (char ch);
int main (void)
{
printf ("Enter a character:\n");
printf ("%d", loca (getchar ()));
}
int loca (char ch)
{
while (ch != EOF)
{
if (((ch >= 'a') && (ch <= 'z')))
{
return ch - 'a' + 1;
}
else if (((ch >= 'A') && (ch <= 'Z')))
{
return ch - 'A' + 1;
}
else
return -1;
}
}
8.
#include <stdio.h>
float power (int a, int b);
int main (void)
{
int index;
int base;
printf ("请输入底数：\n");
scanf ("%d", &base);
printf ("请输入指数：\n");
scanf ("%d", &index);
printf ("结果是：%f", (float) power (base, index));
return 0;
}
float power (int a, int b)
{
float res;
int i;
b = (b < 0) ? b : -b;
if (a == 0)
res = 0;
else if (b == 0)
res = 1;
else
{
res = 1;
for (i = 0; i < -b; i++)
{
res *= a;
}
res = 1 / res;
}
return res;
}
9.
#include <stdio.h>
double power (double, int );
int main (void)
{
double x;
int exp;
printf ("请输入底数和指数：\n");
scanf ("%lf%d", &x, &exp);
printf ("%.3g to the power %d is %.5g\n", x, exp, power (x, exp));
return 0;
}
double power (double n, int p)
{
int i;
double pow=1;
if (p>0)
for (i=1; i<=p; i++)
pow *= n;
else if (p<0)
pow = 1 / power (n, -p);
else if (n != 0)
pow = 1;
else pow = 1 / n;
return pow;
}
10.
#include <stdio.h>
#include <math.h>
float to_base_n (int, int);
int main (void)
{
int base, n;
printf ("Enter base:\n");
scanf ("%d", &base);
printf ("Enter n:\n");
while ((scanf ("%d", &n)) == 1)
{
printf ("The number base to n is: %f\n", to_base_n (base, n));
printf ("Enter base:\n");
scanf ("%d", &base);
printf ("Enter n:\n");
}
return 0;
}
float to_base_n (int base, int n)
{
int a, b;
int i = 0;
float s = 0;
do
{
b = base / n;
a = base % n;
s = a * pow (10, i) + s;
base = b;
i++;
}
while (b != 0);
return s;
}
11.
#include <stdio.h>
int Fib (int num);
int main (void)
{
int num;
char ch;
printf ("请输入你想知道的斐波那契数的序号：\n");
while (scanf ("%u", &num) == 1)
{
if (num <= 0)
{
printf ("输入有误，请重新输入：\n");
continue;
}
printf ("第%d号斐波那契数是：%u\n", num, Fib (num));
printf ("\n请输入你想知道的斐波那契数的序号：\n");
}
}
int Fib (int num)
{
int i, j, a, t;
i = j = 1;
if ((num == 1) || (num == 2))
return 1;
for (a = 2; a < num; a++)
{
t = i + j;
j = i;
i = t;
}
return t;
}