实验3 C语言函数应用编程

task.1

#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;

}

问题1 将学生分数转换为等级，形参类型为int，返回值类型为char.

问题2 无论输入什么数值都会执行line21-28所有语句但是最后结果显示为E

 

task.2

#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;
}

问题1 计算各个位数数字的总和

问题2 能实现同样的输出，一种是迭代方式，一种是递归方式。迭代方式是循环处理，不断累加；递归方式是将问题分解计算

 

task.3

#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=0;

    if(n==0)
        return 1;
    else if(n%2)
        return x*power(x,n-1);
    else{
        t=power(x,n/2);
        return t*t;
    }
}

问题1 计算x的n次方

问题2  if n=0 返回1

　　　if n为奇数 返回x*power(x,n-1)

　　　if n为偶数 返回power(x,n/2)*power(x,n/2)

 

task.4

#include <stdio.h>
int is_prime(int n) {
    int i;
    if (n <= 1) return 0;
    if (n == 2) return 1;
    if (n % 2 == 0) return 0;

    for (i = 3; i * i <= n; i += 2) {
        if (n % i == 0) return 0;
    }
    return 1;
}

int main() {
    int n;
    int count = 0;

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

    printf("\n100以内的孪生素数共有%d个\n", count);

    return 0;
}

 

task.5

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 (m < 0 || m > n) return 0;
    if (m == 0 || m == n) return 1;
    return func(n - 1, m) + func(n - 1, m - 1);
}

递归

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

int func(int n,int m);
int fac(int );

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;
     }
     if(m>n){
        return 0;
     }
     return func(n-1,m)+func(n-1,m-1);
 }

 

 task.6

 

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

int gcd(int,int);

int main(){
    int a,b,c;
    int ans;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF){
        ans=gcd(gcd(a,b),c);
        printf("最大公约数：%d\n\n",ans);
    }
    system("pause");
    return 0;
}

int gcd(int x,int y){
    int r;
    r=x%y;
    while(r!=0){
        x=y;
        y=r;
        r=x%y;
    }
    return y;
}

 

 task.7

#include <stdio.h>
#include <stdlib.h>
void print_charman(int n);

int main() {
    int n;

    printf("Enter n: ");
    scanf("%d", &n);
    print_charman(n); // 函数调用
      system("pause");
    return 0;
}
void print_charman(int n){int i,k,j;
for(i=0;i<n;++i){
    for(j=0;j<i;++j)printf(" \t");
    for(k=0;k<n*2-1-2*i;++k){printf(" O\t");}
    printf("\n");
    for(j=0;j<i;++j)printf(" \t");
    for(k=0;k<n*2-1-2*i;++k){printf("<H>\t");}
    printf("\n");
    for(j=0;j<i;++j)printf(" \t");
    for(k=0;k<n*2-1-2*i;++k){printf("I I\t");}
    printf("\n");}}