实验3

任务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：讲成绩换算成等级，形参类型为整型，返回值类型为字符型

 问题2：所有输出结果都将为E

 

 

 

任务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：算出一个数各位相加的结果

 

问题2：能，一种是迭代的思想，一种是递归的思想

 

任务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;
    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：算出一个数的n次方

 

问题2：是

 

 

任务4

#include <stdio.h>
int is_prime(int n){
    if (n<=2)
    return 0;
    else {for(int i=2;i<n;i++)
    {if(n%i==0)
    return 0;
    }return 1;
    }
}
int main (){printf("100以内的孪生素数\n");
int t=0;
for(int i=3;i<=100;i++) {
    
    if(is_prime(i)&&is_prime(i+2)){
    printf("%d %d\n",i,i+2);
    t++;}
}
    
    printf("共有%d对孪生素数\n",t);
    
    
    
    return 0;
}

任务5

迭代

#include <stdio.h>
int func(int n,int m){
    if (m==0)
    return 1;
    else if (m==n)
    return 1;
    else if (m>n)
    return 0;
    else {
    int a=1;
    int b=1;
    for(int i=1;i<=m;i++)
    {a*=n-i+1;
    b*=m-i+1;    }return a/b;
    }
    }
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;
}

递归

#include <stdio.h>
int func(int n,int m){
    if (m==0)
    return 1;
    else if (m==n)
    return 1;
    else if (m>n)
    return 0;
    else {return func(n-1,m)+func(n-1,m-1);
    
    }
    }
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;
}

 

 任务6

#include<stdio.h>
int gcd(int a,int b,int c)
{int m;
if(a<b)
{m=a;
if (c<a)
m=c;
else m=a;
}
else {m=b;
if(c<b)
m=c;
else m=b;
}for(int i=m;i>=1;i--)
{if (a%i==0&&b%i==0&&c%i==0)
return i;
}
}
int main(){
    int a,b,c;
    int ans;
    while(scanf("%d%d%d",&a,&b,&c)!=EOF){
        ans= gcd(a,b,c);
        printf("最大公约数：%d\n\n",ans);
    }
    return 0;
}

 

 

 

任务7

#include <stdio.h>
#include <stdlib.h>
void print_charman(int n,int m){
    for (int i=m;i>=1;i--)
        printf("\t");
    for (int i=1;i<=2*n-1;i++)printf(" o\t");printf("\n");
    for (int i=m;i>=1;i--)
        printf("\t");
    for (int i=1;i<=2*n-1;i++)printf("<H>\t");printf("\n");
    for (int i=m;i>=1;i--)
        printf("\t");
    for (int i=1;i<=2*n-1;i++)printf("I I\t");printf("\n");
    
}
int main(){
    int n=0;
    printf("input n:");
    scanf("%d",&n);

    
    for (int i=n;i>=1;i--){
        print_charman(i,n-i);
    }
    return 0;
}