实验3

task1

源代码

#include<stdio.h>

char score_to_grade(int score);

int main(){
int score;
char grade;

while(scanf("%d",&score)!=EOF){
grade=score_to_grade(score);
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;
}

运行结果

问题1：定义一个函数，字符型，字母

问题2：有问题，程序会一直进行下去

task2

源代码

#include<stdio.h>

int sum_digits(int n);

int main(){
int n;
int ans;

while(printf("Enter n:"),scanf("%d",&n)!=EOF){
ans=sum_digits(n);
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;
}

运行结果

问题一：定义一个新函数

问题二：能，改为非递归函数也可以

task3

源代码

#include<stdio.h>

int power(int x,int n);

int main(){
int x,n;
int ans;

while(printf("Enter x and n:"),scanf("%d%d", &x, &n)!=EOF){
ans = power(x,n);
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;
}
}

运行结果

问题1：输出一个数的幂和他的次方值

问题2：如果n=0，返回一，如果n是偶数，则运行程序

task4

源代码

#include <stdio.h>

int is_prime(int n) {
    if (n <= 1) {
        return 0;
    }
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return 0;
        }
    }
    return 1;
}

int main() {
    int count = 0; 

    printf("100以内的孪生素数有：\n");
    for (int n = 1; n <= 98; n++) {
        if (is_prime(n) && is_prime(n + 2)) {
            printf("(%d, %d)\n", n, n + 2);
            count++;
        }
    }

    printf("孪生素数的总数为：%d\n", count);

    return 0;
}

运行结果

task5

源代码

#include <stdio.h>

void hanoi(int n, char from, char to, char aux, int* moveCount) {
    if (n == 1) {
        printf(" 1 from %c --> %c\n", from, to);
        (*moveCount)++;
    } else {
        hanoi(n - 1, from, aux, to, moveCount);
        printf(" %d from %c --> %c\n", n, from, to);
        (*moveCount)++;
        hanoi(n - 1, aux, to, from, moveCount);
    }
}

int main() {
    int  n;

    for (int i = 0; i < 100; i++) {
        printf("Enter the number of disks for case %d: ", i + 1);
        scanf("%d", &n);

        int moveCount = 0;

        printf("Movement plan for %d disks:\n", n);
        hanoi(n, 'A', 'C', 'B', &moveCount);

        printf("一共移动了: %d\n\n", moveCount);
    }

    return 0;
}

运行结果

task6

源代码 迭代

#include <stdio.h>
int func(int n, int m); 
int main() {
int n, m;
int ans;
while(scanf("%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(n<m) {
        return 0;
    }
    int C[n + 1][m + 1];
    
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= (i < m ? i : m); j++) {
            if (j == 0 || j == i) {
                C[i][j] = 1;
            } else {
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            }
        }
    }
    
    return C[n][m];
}

运行结果

源代码递归

#include <stdio.h>
int func(int n, int m);
int main() {
int n, m;
int ans;
while(scanf("%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 == 0 || m == n) {
        return 1;
    } 
    else if(n<m){
        return 0;
    }
    else {
        return func(n - 1, m - 1) + func(n - 1, m);
    }
}

运行结果

task7

源代码

#include<stdio.h>
#include <stdlib.h>

int print_charman(int n);

int main() {
    int n;

    printf("Enter n: ");
    scanf("%d", &n);
    print_charman(n);

    return 0;
}

int print_charman(int n){
     int x = 0,i,j;
    for (i = n; i >= 1; i--)
     {
         for (j = 0; j <= x; j++)
         {
             printf("\t");
         }
         for (j = 0; j < 2 * i - 1; j++)
         {
             printf(" O\t");
         }
         printf("\n");
         for (j = 0; j <= x; j++)
         {
             printf("\t");
         }
         for (j = 0; j < 2 * i - 1; j++)
         {
             printf("<H>\t");
         }
         printf("\n");
         for (j = 0; j <= x; j++)
         {
             printf("\t");
         }
         for (j = 0; j < 2 * i - 1; j++)
         {
             printf("I I\t");
         }
         printf("\n");
         x++;
     }

         }

运行结果