实验三

任务1

task1.1 源代码

#include<stdio.h>
long long fac(int n);
int main()
{
    int i, n;
    printf("Enter n: "); 
    scanf_s("%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;
}

修改过一行的代码

#include<stdio.h>
long long fac(int n);
int main()
{
    int i, n;
    printf("Enter n: "); 
    scanf_s("%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;
}

运行效果

 

task 1.2

#include<stdio.h>
int func(int, int);
int main()
{
    int k = 4, m = 1, p1, p2;
    p1 = func(k, m); // 函数调用 
    p2 = func(k, m);// 函数调用 
    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变量不会随着函数的结束而被释放

static变量不会被再次赋初值

 

任务2

迭代版

#include<stdio.h>
void printSymbol(int n, char symbol)
{
    for(int i=0;i<n;i++)
    {
        printf("%c", symbol);
    }
}
int main()
{
    int n;
    char symbol;
    while (scanf_s("%d %c", &n, &symbol) != EOF)
    {
        printSymbol(n, symbol); // 函数调用 
        printf("\n");
    }
    return 0;
}

递归版

#include<stdio.h>
void printSymbol(int n, char symbol)
{
    if (n == 0)return;
    printSymbol(n - 1, symbol);
    printf("%c", symbol);
}
int main()
{
    int n;
    char symbol;
    while (scanf_s("%d %c", &n, &symbol) != EOF)
    {
        printSymbol(n, symbol); // 函数调用 
        printf("\n");
    }
    return 0;
}

 我认为此题迭代更好

因为迭代的思维方式在此题中更加直观易懂,且代码量也和递归方式相差不大

 

任务3

#include<stdio.h>
long long fun(int n)
{
    if (n == 0)return 0;
    else
    {
        return 2 * fun(n - 1)+1;
    }
}
int main()
{
    int n; 
    long long f; 
    while (scanf_s("%d", &n) != EOF)
    {
        f = fun(n); // 函数调用
        printf("n = %d, f = %lld\n", n, f);
    }
    return 0;
}

 

任务4

#include<stdio.h>
#include<math.h>
int isPrime(int n)
{
    for(int i=2;i<=sqrt(n);i++)
    {
        if (n % i == 0)return 0;
    }
    return 1;
}
int main()
{
    int sum = 0;
    for(int i=101;i<=200;i++)
    {
        if(!isPrime(i))
        {
            printf("%6d", i);
            sum++;
            if (sum % 10 == 0)printf("\n");
        }
    }
    printf("\n一共有%d个非素数", sum);
    return 0;
}

 

 任务5

#include<stdio.h>
#include<math.h>
long fun(long s)
{
    long a=0, b=1,num=0;//a用来存放每位的数,b用来记位数
    while(s>0)
    {
        a = s % 10;
        if(a%2!=0)
        {
            num += a * b;
            b *= 10;
        }
        s /= 10;
    }
    return num;
}
int main()
{
    long s, t; 
    printf("Enter a number: "); 
    while (scanf_s("%ld", &s) != EOF) 
    {
        t = fun(s); // 函数调用
        printf("new number is: %ld\n\n", t); 
        printf("Enter a number: "); 
    }
    return 0;
}

 

任务6

#include<stdio.h>
#include<math.h>
double jie_cheng(int x)
{
    if (x == 1)return 1;
    else
    {
        return x * jie_cheng(x - 1);
    }
}
double fun(int n)
{
    double sum=0;
    for(int i=1;i<=n;i++)
    {
        if (i % 2 == 0)sum -= (1 / jie_cheng(i));
        else sum += (1 / jie_cheng(i));
    }
    return sum;
}
int main()
{
    int n; 
    double s;
    printf("Enter n(1~10): ");
    while (scanf_s("%d", &n) != EOF) 
    {
        s = fun(n); // 函数调用
        printf("n = %d, s= %f\n\n", n, s);
        printf("Enter n(1~10): "); 
    }
    return 0;
}

 

 

posted on 2021-11-24 11:09  彭涌涛  阅读(26)  评论(1编辑  收藏  举报

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