C语言程序设计题目（急）

// 不使用递归的 f 函数

double f(double x, int n) {

double result = 0

int sign = 1

for (int i = 0i <ni++) {

result += sign * x

x *= x

sign = -sign

}

return result

}

// 使用递归的 Rf 函数

double Rf(double x, int n) {

if (n == 0) {

return 0

}

return x - Rf(x * x, n - 1)

}

int main() {

double x

int n

for (int i = 0i <3i++) {

printf("请输入 x 和 n 的值：")

scanf("%lf%d", &x, &n)

printf("f(x, n) = %lf\n", f(x, n))

printf("Rf(x, n) = %lf\n", Rf(x, n))

}

return 0

}

这段程序中，f 函数使用了一个循环来计算函数值，而 Rf 函数使用了递归。 main 函数提示用户输入 x 和 n 的值，并调用 f 和 Rf 函数计算函数值，最后将结果输出到屏幕上。

#include "stdio.h"

void main(void)

{

int n=5,m=10,i=1

long sum=1

for(i<=ni++)

{

sum*=i

}

printf("\n5!=%d",sum)

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

{

sum*=i

}

printf("\n10!=%d",sum)

}

题目2

#include "stdio.h"

#include "string.h"

struct Student

{

char s_Name[25]

long n_Code

int n_English

int n_Math

int n_Computer

}student_1,student_2

void main(void)

{

printf("\nStudent1:\nName:")

scanf("%s",&student_1.s_Name)

printf("StudentNum:")

scanf("%d",&student_1.n_Code)

printf("English Score:")

scanf("%d",&student_1.n_English)

printf("Math Score:")

scanf("%d",&student_1.n_Math)

printf("Computer Score:")

scanf("%d",&student_1.n_Computer)

printf("\nStudent2:\nName:")

scanf("%s",&student_2.s_Name)

printf("StudentNum:")

scanf("%d",&student_2.n_Code)

printf("English Score:")

scanf("%d",&student_2.n_English)

printf("Math Score:")

scanf("%d",&student_2.n_Math)

printf("Computer Score:")

scanf("%d",&student_2.n_Computer)

printf("\nStudent1:\nName:%s\nStudent Number:%d\nEnglish Score:%d\nMath Score:%d\nComputer Score:%d\n",student_1.s_Name,student_1.n_Code,student_1.n_English,student_1.n_Math,student_1.n_Computer)

printf("\nStudent2:\nName:%s\nStudent Number:%d\nEnglish Score:%d\nMath Score:%d\nComputer Score:%d\n",student_2.s_Name,student_2.n_Code,student_2.n_English,student_2.n_Math,student_2.n_Computer)

}