实验3作业

第一题

include <stdio.h>

char score_to_grade(int score); // 函数声明
int main() {
int score;
char grade;
while(scanf("%d", &score) != EOF) {
grade = score_to_grade(score); // 函数调用
printf("分数: %d, 等级: %c\n\n", score, grade);
} return 0;
}
char score_to_grade(int score) {
char ans;
switch(score/10) {
case 10:
case 9: ans = 'A'; break;
case 8: ans = 'B'; break;
case 7: ans = 'C'; break;
case 6: ans = 'D'; break;
default: ans = 'E';
}
return ans;
}
结果是
A
B
A
C
E
返回值是等级
这样的话会同时出现ABCDE

第二提

include <stdio.h>

int sum_digits(int n); // 函数声明
int main() {
int n;
int ans;
while(printf("Enter n: "), scanf("%d", &n) != EOF) {
ans = sum_digits(n); // 函数调用
printf("n = %d, ans = %d\n\n", n, ans);
} return 0;
}
int sum_digits(int n) {
int ans = 0;
while(n != 0) {
ans += n % 10;
n /= 10;
} return ans;
}
作用是计算各分位上数字之和
可以。我乍一看没问题
第三题

include <stdio.h>

int power(int x, int n); // 函数声明
int main() {
int x, n;
int ans;
while(printf("Enter x and n: "), scanf("%d%d", &x, &n) != EOF) {
ans = power(x, n); // 函数调用
printf("n = %d, ans = %d\n\n", n, ans);
} return 0;
}
int power(int x, int n) {
int t;
if(n == 0)
return 1;
else if(n % 2)
return x * power(x, n-1);
else {
t = power(x, n/2);
return t*t;
}
}
power的作用是计算次方，如3的5次方

第四题

include <stdio.h>

include <math.h>

int sushu(int n) {
if (n < 2) {
return 0;
}
for (int i = 2; i <= sqrt(n); i++) {
if (n % i == 0) {
return 0;
}
}
return 1;
}
int main() {
int s = 0;
printf("素数:\n");
for (int i = 2; i < 99; i++) {
if (sushu(i) && sushu(i + 2)) {
printf("%d, %d\n", i, i + 2);
s++;
}
}
printf("总数: %d\n", s);
return 0;
}

第五题

include<stdio.h>

include<stdlib.h>

int count = 0;
void hanoi(unsigned int n, char from, char temp, char to);
void moveplate(unsigned int n, char from, char to);
int main()
{
unsigned int n;
while (scanf_s("%u", &n) != EOF)
{
count = 0;
hanoi(n, 'A', 'B', 'C');
printf("一共移动的次数为：%d", count);
}
system("pause");
return 0;
}
void hanoi(unsigned int n, char from, char temp, char to)
{
if (n == 1)
moveplate(n, from, to);
else
{
hanoi(n - 1, from, to, temp);
moveplate(n, from, to);
hanoi(n - 1, temp, from, to);
}
}
void moveplate(unsigned int n, char from, char to)
{
printf("%u:%c-->%c\n", n, from, to);
count++;
}

第六题

include <stdio.h>

int func(int n, int m);
int main() {
int n, m;
int ans;
while (scanf_s("%d%d", &n, &m) != EOF) {
ans = func(n, m);
printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);
}
return 0;
}
int func(int n, int m) {
if (m > n) {
return 0;
}
int s = 1;
for (int i = n; i > n - m; i = i - 1){
s = s * i;
}
for (int i = 1; i <= m; i = i + 1) {
s /= i;
}
return s;
}
上面是第一种方法。

include <stdio.h>

int func(int n, int m);
int main() {
int n, m;
int ans;
while (scanf_s("%d%d", &n, &m) != EOF) {
ans = func(n, m);
printf("n = %d, m = %d, ans = %d\n\n", n, m, ans);
}
return 0;
}
int func(int n, int m) {
if(m0||mn){
return 1;
}
if (m > n) {
return 0;
}
return func(n - 1, m) + func(n - 1, m - 1);
}
第二种。

第七题

include <stdio.h>

int gcd(int a, int b, int c);
int main() {
int a, b, c;
int ans;
while (scanf_s("%d%d%d", &a, &b, &c) != EOF) {
ans = gcd(a, b, c);
printf("最大公约数: %d\n\n", ans);
}
return 0;
}
int gcd(int a, int b, int c) {
int min = a;
if (b < min)
min = b;
if (c < min)
min = c;
for(int i=min;i>0;i=i-1){
if (a % i == 0 && b % i == 0 && c % i == 0) {
return i;
}
}

}