第五章
注：以下所有代码需要添加头文件<stdio.h>、<math.h>
- 例5.6的三个程序
代码如下
void matrix_4x5_program1();
void matrix_4x5_program2();
void matrix_4x5_program3();
int main()
{
matrix_4x5_program1();
matrix_4x5_program2();
matrix_4x5_program3();
return 0;
}
void matrix_4x5_program1()
{
int num = 0;
for(int i = 1; i <= 4; i++)
{
for(int j = 1; j <= 5; j++)
{
num++;
printf("%d\t", i * j);
if(num % 5 == 0) printf("\n");
}
}
printf("\n");
}
void matrix_4x5_program2()
{
int num = 0;
for(int i = 1; i <= 4; i++)
{
for(int j = 1; j <= 5; j++)
{
num++;
if(i == 3)
{
printf("\n");
num--;
break;
}
printf("%d\t", i * j);
if(num % 5 == 0) printf("\n");
}
}
printf("\n");
}
void matrix_4x5_program3()
{
int num = 0;
for(int i = 1; i <= 4; i++)
{
for(int j = 1; j <= 5; j++)
{
num++;
if(i == 3 && j == 1) continue;
printf("%d\t", i * j);
if(num % 5 == 0) printf("\n");
}
}
}
- 补充例5.7程序
代码如下
int main()
{
int sign = 1, n = 1, count = 0;
double pi = 0.0, term = 1.0;
while(fabs(term) > 1e-8)
{
pi += term;
n += 2;
sign *= -1;
term = sign * 1.0 / n;
count++;
}
printf("pi = %10.8f\n", pi * 4);
printf("count = %d\n", count);
return 0;
}
- 求两个正整数的最大公约数和最小公倍数
代码如下
void gcd_lcm();
int main()
{
gcd_lcm();
return 0;
}
void gcd_lcm()
{
int a, b, gcd, lcm, temp;
printf("a = ");
scanf("%d", &a);
printf("b = ");
scanf("%d", &b);
if(a < b)
{
temp = a;
a = b;
b = temp;
}
lcm = a * b;
while(b != 0)
{
gcd = a % b;
a = b;
b = gcd;
}
printf("gcd = %d, lcm = %d\n", a, lcm / a);
}
- 统计一行字符中的英文字母、数字、空格和其他字符的个数
代码如下
void count_char();
int main()
{
count_char();
return 0;
}
void count_char()
{
char ch;
int letters = 0, space = 0, digit = 0, other = 0;
while((ch = getchar()) != '\n')
{
if((ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z'))
{
letters++;
}
else if(ch == ' ')
{
space++;
}
else if(ch >= '0' && ch <= '9')
{
digit++;
}
else
{
other++;
}
}
printf("letters = %d, space = %d, digit = %d, other = %d\n", letters, space, digit, other);
}
- 求数列的前n项和
代码如下
void sum_sequences();
int main()
{
sum_sequences();
return 0;
}
void sum_sequences()
{
int a, len, temp = 0, sum = 0;
printf("a = ");
scanf("%d", &a);
printf("len = ");
scanf("%d", &len);
for(int i = 1; i <= len; i++)
{
temp += a;
sum += temp;
a *= 10;
}
printf("sum = %d\n", sum);
}
- 求阶乘数列前n项的和
代码如下
void factorial_sequences();
int main()
{
factorial_sequences();
return 0;
}
void factorial_sequences()
{
int len = 20;
double temp =1, sum = 0;
for(int i = 1; i <= len; i++)
{
temp *= i;
sum += temp;
}
printf("sum = %22.15e\n", sum);
}
- 求数列的前n项和
代码如下
void sum_sequences();
int main()
{
sum_sequences();
return 0;
}
void sum_sequences()
{
int len1 = 100, len2 = 50, len3 = 10;
double sum1 = 0, sum2 = 0, sum3 = 0;
for(int i = 1; i <= len1; i++)
{
sum1 += i;
}
for(int i = 1; i <= len2; i++)
{
sum2 += pow(i, 2);
}
for(int i = 1; i <= len3; i++)
{
sum3 += pow(i, -1);
}
printf("sum = %15.6f", sum1 + sum2 + sum3);
}
- 输出1000以内所有的水仙花数
代码如下
/*
水仙花数：153 = 1^3 + 5^3 + 3^3
- 定义三个整型变量，分别存储百位，十位，个位
- 计算百位直接除以100，hundred = 123 / 100 = 1
- 计算十位直接除以10，ten = 123 / 10 - hundred * 10 = 2
- 计算个位直接取余，one = 123 % 10 = 3
- 定义一个变量，存储百位的三次方 + 十位的三次方 + 个位的三次方
- 判断当前数是否等于这个变量
*/
void is_shuixianhua();
int main()
{
is_shuixianhua();
return 0;
}
void is_shuixianhua()
{
int num, i, j, max = 1000, min = 100;
printf("parcissus numbers are: \n");
for(min; min < max; min++)
{
i = min / 100;
j = min / 10 - i * 10;
num = min % 10;
if(min == i * i * i + j * j * j + num * num * num)
{
printf("%d ", min);
}
}
printf("\n");
}
- 输出1000以内所有的完美数
代码如下
/*
从2开始，因为2的因子只有1和它本身，所以从2开始，有以下变量
- a，从2开始
- sum，因子的累加和，验证a == sum
- i，计算因子，满足a % i == 0
*/
#define MAX 1000
void is_perfect_program1();
void is_perfect_program2();
int main()
{
is_perfect_program1();
is_perfect_program2();
return 0;
}
void is_perfect_program1()
{
int m, s;
for(m = 2; m < MAX; m++)
{
// 从2开始，因为2的因子只有1和它本身，所以从2开始
s = 0;
for(int i = 1; i < m; i++)
{
// 计算m的因子，将因子累加
if(m % i == 0) s += i;
}
if(s == m)
{
// 如果因子的累加和等于m，那么m就是完美数
printf("%d, its factor is: ", m);
for(int i = 1; i < m; i++)
{
// 再次计算m的因子，并将其输出
if(m % i == 0) printf("%d ", i);
}
printf("\n");
}
}
}
void is_perfect_program2()
{
int k1, k2, k3, k4, k5, k6, k7, k8, k9, k10;
int a = 2, sum = 0, count;
for(a; a <= MAX; a++)
{
count = 0;
sum = a; // 存放尚未求出的因子之和，先存入本身
for(int i = 1; i < a; i++) // 计算因子
{
if(a % i == 0)
{
count++; // 计算因子的个数
sum -= i; // 减去所找到的因子
switch(count) // 将找到的因子赋值给对应的变量
{
case 1:
k1 = i;
break;
case 2:
k2 = i;
break;
case 3:
k3 = i;
break;
case 4:
k4 = i;
break;
case 5:
k5 = i;
break;
case 6:
k6 = i;
break;
case 7:
k7 = i;
break;
case 8:
k8 = i;
break;
case 9:
k9 = i;
break;
case 10:
k10 = i;
break;
}
}
}
if(sum == 0)
{
printf_s("%d, its factors are ", a);
if(count > 1) printf_s("%d, %d", k1, k2); // 数至少2个因子时，输出前2个因子，即1和它本身
if(count > 2) printf_s(", %d", k3);
if(count > 3) printf_s(", %d", k4);
if(count > 4) printf_s(", %d", k5);
if(count > 5) printf_s(", %d", k6);
if(count > 6) printf_s(", %d", k7);
if(count > 7) printf_s(", %d", k8);
if(count > 8) printf_s(", %d", k9);
if(count > 9) printf_s(", %d", k10);
printf_s("\n");
}
}
}
- 计算分数数列前n项和
代码如下
/*
第一个分数是2/1，后一个分数分子是前一个分数分子分母之和，分母是前一个分数分子
- 定义两个变量，分子和分母；一个临时变量，存储分子；一个变量，存储分数的和
- 在每一次for循环中，先将当前分数累加至和变量中；再将当前分数的分子赋值给临时变量；
计算下一个分数的分子和分母，进行下一次循环
*/
void sum_fractional_sequences_program1();
void sum_fractional_sequences_program2();
int main()
{
sum_fractional_sequences_program1();
sum_fractional_sequences_program2();
return 0;
}
void sum_fractional_sequences_program1()
{
int numerator1 = 2, denominator1 = 1, numerator2, denominator2;
double sum = 0, fraction1, fraction2;
fraction1 = (double)numerator1 / denominator1;
sum += fraction1;
for(int i = 1; i < 20; i++)
{
numerator2 = numerator1 + denominator1;
denominator2 = numerator1;
fraction2 = (double)numerator2 / denominator2;
sum += fraction2;
fraction1 = fraction2;
numerator1 = numerator2;
denominator1 = denominator2;
}
printf("sum = %f\n", sum);
}
void sum_fractional_sequences_program2()
{
int numerator = 2, denominator = 1, temp;
double sum = 0;
for(int i = 1; i <= 20; i++)
{
// 对program1的改进，减少变量的冗余定义
sum += (double)numerator / denominator;
temp = numerator;
numerator = numerator + denominator;
denominator = temp;
}
printf("sum = %f\n", sum);
}
- 计算球所经过的高度以及第10次反弹的高度
代码如下
void ball_rebound_height_program1();
int main()
{
ball_rebound_height_program1();
return 0;
}
void ball_rebound_height_program1()
{
double height = 100, sum = 0;
sum += height;
for(int i = 1; i < 10; i++)
{
height /= 2;
sum += height * 2;
}
printf("sum = %f\n", sum);
printf("height = %f\n", height);
}
- 猴子吃桃问题
代码如下
/*
数学分析，quantities(n) = (quantities(n - 1) + 1) * 2， n > 1
*/
void peach_quantity_program1();
int peach_quantity_program2(int len);
int main()
{
peach_quantity_program1();
printf("peach = %d\n", peach_quantity_program2(10));
return 0;
}
void peach_quantity_program1()
{
int peach = 1;
for(int i = 1; i < 10; i++)
{
peach = (peach + 1) * 2;
}
printf("peach = %d\n", peach);
}
int peach_quantity_program2(int len)
{
// 递归
int result = 0;
if(len == 1) result = 1;
else result = (peach_quantity_program2(len - 1) + 1) * 2;
return result;
}
- 迭代法求平方根
代码如下
/*
迭代法，求平方根
- 设定一个初始值x0
- 设定一个临时变量xn，用迭代公式计算xn+1
- 将xn赋值给x0，xn+1赋值给xn，进行下一次循环
- 直到|xn - x0| < 1e-5
float sqrt_root(float num);
int main()
{
printf("sqrt = %f\n", sqrt_root(2));
return 0;
}
float sqrt_root(float num)
{
float x0, xn;
x0 = num / 2;
xn = (x0 + num / x0) / 2;
while(fabs(xn - x0) > 1e-5)
{
x0 = xn;
xn = (x0 + num / x0) / 2;
}
return xn;
}
- 牛顿法求方程根
代码如下
/*
牛顿迭代法即牛顿切线法
- 设定一个初始值x0，f(x0) = f'(x0)(x1 - x0)
- 将x0赋值给x1，x1赋值给x0，进行下一次循环
- 直到|xn - x0| < 1e-5
*/
double solve_quadratic_equation();
int main()
{
double root0 = solve_quadratic_equation();
printf_s("root is %5.2f\n", root0);
return 0;
}
double solve_quadratic_equation()
{
double x1, x0, f, f1;
x1 = 1.5;
do
{
x0 = x1;
f = ((2 * x0 - 4) * x0 + 3) * x0 - 6;
f1 = (6 * x0 - 8) * x0 + 3;
x1 = x0 - f / f1;
} while (fabs(x1 - x0) > 1e-5);
return x1;
}
- 二分法求方程根
代码如下
/*
二分法，求方程的根
- 设定一个初始值x1和x2，f(x1)和f(x2)，若f(x1)和f(x2)同号，怎重新设定x1和x2直至f(x1)和f(x2)异号
- 求出x0 = (x1 + x2) / 2，f(x0)
- 若f(x0)和f(x1)同号，则在[x0, x2]中寻根，用x0代替x1，f(x0)代替f(x1)
- 若f(x0)和f(x1)同号，则在[x1, x0]中寻根，用x0代替x2，f(x0)代替f(x2)
- 直到|f(x0)| < 1e-5
*/
void binary_search();
int main()
{
binary_search();
return 0;
}
void binary_search()
{
float x0, x1 = -10, x2 = 10, fx0, fx1, fx2;
do
{
fx1 = x1 * ((2 * x1 - 4) * x1 + 3) - 6;
fx2 = x2 * ((2 * x2 - 4) * x2 + 3) - 6;
} while (fx1 * fx2 > 0);
do
{
x0 = (x1 + x2) / 2;
fx0 = x0 * ((2 * x0 - 4) * x0 + 3) - 6;
if((fx0 * fx1) < 0)
{
x2 = x0;
fx2 = fx0;
}
else
{
x1 = x0;
fx1 = fx0;
}
} while (fabs(fx0) > 1e-5);
printf_s("root is %5.2f\n", x0);
}
- 输出图案
代码如下
/*
打印图案一半分为上下、左右来进行打印，要理清逻辑
*/
void print_pattern();
int main()
{
print_pattern();
return 0;
}
void print_pattern()
{
for(int i = 0; i <= 3; i++)
{
for(int j = 0; j <= 2 - i; j++)
{
printf(" ");
}
for(int k = 0; k <= 2 * i; k++)
{
printf("*");
}
printf("\n");
}
for(int i = 0; i <= 2; i++)
{
for(int j = 0; j <= i; j++)
{
printf(" ");
}
for(int k = 0; k <= 4 - 2 * i; k++)
{
printf("*");
}
printf("\n");
}
}
- 编程找出三对赛手的名单
代码如下
void print_name();
int main()
{
print_name();
return 0;
}
void print_name()
{
char i, j, k;
for(i = 'x'; i <= 'z'; i++)
{
for(j = 'x'; j <= 'z'; j++)
{
if(i != j)
{
for(k = 'x'; k <= 'z'; k++)
{
if(i != k && j != k)
{
if(i != 'x' && k != 'x' && k != 'z')
{
printf("A---%c\nB---%c\nC---%c\n", i, j, k);
}
}
}
}
}
}
}
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