实验三

实验一

task1-1（填代码前）

#include <stdio.h>
long long fac(int n); 
int main()
{
    int i, n;
    printf("Enter n: ");
    scanf("%d", &n);
    for (i = 1; i <= n; ++i)
    printf("%d! = %lld\n", i, fac(i));
    return 0;
}

long long fac(int n)
{
    static long long p = 1;
    p = p * n;
    return p;
}

 

 

task1-1（增加代码后）

#include <stdio.h>
long long fac(int n); 
int main()
{
    int i, n;
    printf("Enter n: ");
    scanf("%d", &n);
    for (i = 1; i <= n; ++i)
    printf("%d! = %lld\n", i, fac(i));
    return 0;
}

long long fac(int n)
{
    static long long p = 1;
    printf("p = %lld\n", p);
    p = p * n;
    return p;
}

task 1-2

#include <stdio.h>
int func(int, int); 
int main()
{
    int k = 4, m = 1, p1, p2;
    p1 = func(k, m); 
    p2 = func(k, m); 
    printf("%d,%d\n", p1, p2);
    return 0;
}

int func(int a, int b)
{
    static int m = 0, i = 2;
    i += m + 1;
    m = i + a + b;
    return (m);
}

 

 static变量为静态变量，即再次进入函数时，将保存上次运行的结果。

实验二
task 2-1

#include <stdio.h>
void printSymbol(int n, char symbol); 
int main()
{
    int n;
    char symbol;
    while( scanf("%d %c", &n, &symbol) != EOF )
    {
        printSymbol(n, symbol); 
        printf("\n");
    }
    return 0;
}

void printSymbol(int n, char symbol){
    int i;
    for(i=0;i<n;i++){
        printf("%c",symbol);
    }
}

 

 

task 2-2

#include <stdio.h>
void printSymbol(int n, char symbol); 
int main()
{
    int n;
    char symbol;
    while( scanf("%d %c", &n, &symbol) != EOF )
    {
        printSymbol(n, symbol); 
        printf("\n");
    }
    return 0;
}

void printSymbol(int n, char symbol){
    if(n==1){
        printf("%c",symbol);
    }else{
        printSymbol(n-1,symbol);
        printf("%c",symbol);
    }
}

在这个问题中，迭代更好。不需像递归那样不断调用函数，使得代码运行过程中，效率更高。而且算法简单直观。

实验三

#include <stdio.h>
long long fun(int n); 
long long mypow(int n);

int main()
{
    int n;
    long long f;
    while (scanf("%d", &n) != EOF)
    {
        f = fun(n); 
        printf("n = %d, f = %lld\n", n, f);
    }
    return 0;
}

long long fun(int n){
    int result;
    result=mypow(n)-1;
    return result;
}
long long mypow(int n){
    int result=1;
    if(n==1){
        result*=2;
    }else{
        result=2*mypow(n-1);
    }
    return result;
}

 

实验四


#include <stdio.h>
int isPrime(int n);

int main(){
    int i,k=0,n=0;
    
    for(i=101;i<=200;i++){
        if(isPrime(i)==0){
            printf("%5d",i);
            k++;
            n++;
            if(k%10==0){
                printf("\n");
                k=0;
            } 
        }
    }
    printf("\n");
    printf("一共有%d个非素数\n",n);
    
    return 0;
}

int isPrime(int n){
    int j;
    for(j=2;j<n;j++){
        if(n%j==0){
            break;
        }    
    }
    if(j<n){
        return 0;
    }else{
        return 1;
    }
}

 

实验五

#include <stdio.h>
#include<math.h>
long fun(long s); 

int main()
{
    long s, t;
    printf("Enter a number: ");
    while (scanf("%ld", &s) != EOF)
    {
        t = fun(s); 
        printf("new number is: %ld\n\n", t);
        printf("Enter a number: ");
    }
    return 0;
}

long fun(long s){
    int a,i=0;
    long sum=0;
    while(s>0){
        a=s%10;
        if(a%2!=0){
            sum+=a*pow(10,i);
            i++;
        }
        s/=10;
    }
    return sum;
}

 

 

 

 

实验六

#include <stdio.h>
#include<math.h>
double fun(int n);
int factorial(int n);

int main()
{
    int n;
    double s;
    printf("Enter n(1~10): ");
    while (scanf("%d", &n) != EOF)
    {
        s = fun(n); 
        printf("n = %d, s= %f\n\n", n, s);
        printf("Enter n(1~10): ");
    }
    return 0;
}

double fun(int n){
    double sum=0;
    if(n==1){
        sum=1;
    }else{
        sum=pow(-1,n-1)/factorial(n)+fun(n-1);
    }
    return sum;
}

int factorial(int n){
    int fac=0;
    if(n==1){
        fac=1;
    }else{
        fac=n*factorial(n-1);
    }
    return fac;
}

 

 

 

实验收获：

1.函数的定义、调用等有更加深入的理解

2.区别递归和迭代

3.对于写题有更多的经验（注意细节、格式、题意等等）