PTA函数部分20题全部答案

//6-1 使用函数计算两点间的距离

#include 
#include 

double dist(double x1, double y1, double x2, double y2);

int main()
{
    double x1, y1, x2, y2;

    scanf("%lf %lf %lf %lf", &x1, &y1, &x2, &y2);
    printf("dist = %.2f\n", dist(x1, y1, x2, y2));

    return 0;
}

double dist(double x1, double y1, double x2, double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

//6-2 符号函数2

#include 

int sign(int x);

int main()
{
    int x;

    scanf("%d", &x);
    printf("sign(%d) = %d\n", x, sign(x));

    return 0;
}

int sign(int x)
{
    if (x == 0)
        return 0;
	return (x > 0) ? 1 : -1;
}

//6-3 使用函数求Fibonacci数
 

#include 

int fib(int n);

int main()
{
    int n;

    scanf("%d", &n);
    printf("%d\n", fib(n));

    return 0;
}

1.递归，但会超时

Vint fib(int n)
{
    if (n == 1 || n == 2)
        return 1;
    else
        return fib(n - 1) + fib(n - 2);
}

2.一般方法

int fib(int n)
{
    if (n == 1 || n == 2)
     return 1;
    int a = 1, b = 1;
    int tmp = 0;
    for (int i = 3; i <= n; i++)
    {
        tmp = b;
        b = a + b;
        a = tmp;
    }
    return b;
}

//6-4 输出每个月的天数
 

#include 

int MonthDays(int year, int month);

int main()
{
    int ndays, month, year;

    scanf("%d", &year);
    for (month = 1; month <= 12; month++) {
        ndays = MonthDays(year, month);
        printf("%d ", ndays);
    }

    return 0;
}

int MonthDays(int year, int month)
{
    int flag = 1;
    if (year % 4 == 0 && year % 100 != 0)
        flag = 0;
    if (year % 400 == 0)
        flag = 0;
    switch (month)
    {
    case 1:return 31;
    case 2:if (flag) return 28; else return 29;
    case 3:return 31;
    case 4:return 30;
    case 5:return 31;
    case 6:return 30;
    case 7:return 31;
    case 8:return 31;
    case 9:return 30;
    case 10:return 31;
    case 11:return 30;
    case 12:return 31;
    }
}

//6-5 使用函数求素数和
 

#include 
#include 

int prime(int p);
int PrimeSum(int m, int n);

int main()
{
	int m, n, p;

	scanf("%d %d", &m, &n);
	printf("Sum of ( ");
	for (p = m; p <= n; p++) {
		if (prime(p) != 0)
			printf("%d ", p);
	}
	printf(") = %d\n", PrimeSum(m, n));

	return 0;
}

1.

int prime(int n)
{
	int flag = 1;
	if (n <= 1)
		return 0;
	for (int i = 2; i < n; i++)
	{
		if (n % i == 0)
		{
			flag = 0;
			break;
		}
	}
	if (flag == 1)
		return n;
	else
		return 0;
}

2.

int PrimeSum(int m, int n)
{
	int sum = 0;
	for (int ret = 0, j = m; j <= n; j++)
	{
		ret = prime(j);
		if (ret)
		{
			sum += ret;
		}
	}
	return sum;
}

//6-6 字符金字塔

#include 

void CharPyramid(int n, char ch);

int main()
{
    int n;
    char ch;

    scanf("%d %c", &n, &ch);
    CharPyramid(n, ch);

    return 0;
}

void CharPyramid(int n, char ch) {
    int i, j, k;
    for (i = 1; i <= n; i++) {
        for (j = 1; j <= n - i; j++) {
            printf(" ");
        }for (k = 1; k <= i; k++) {
            printf("%c ", ch);
        }printf("\n");
    }
}

//6-7 使用函数计算两个复数之积
 

#include 

double result_real, result_imag;
void complex_prod(double x1, double y1, double x2, double y2);

int main(void)
{
    double imag1, imag2, real1, real2;

    scanf("%lf %lf", &real1, &imag1);
    scanf("%lf %lf", &real2, &imag2);
    complex_prod(real1, imag1, real2, imag2);
    printf("product of complex is (%f)+(%f)i\n", result_real, result_imag);

    return 0;
}

void complex_prod(double x1, double y1, double x2, double y2)
{
    result_real = (x1 * x2 - y1 * y2);
    result_imag = (x1 * y2 + x2 * y1);
}

//6-8 使用函数求余弦函数的近似值
 

//思路一定要清晰
 

#include 
#include 

double funcos(double e, double x);

int main()
{
    double e, x;

    scanf("%lf %lf", &e, &x);
    printf("cos(%.2f) = %.6f\n", x, funcos(e, x));

    return 0;
}

double funcos(double e, double x)
{
    double res = 1, nn = 2,tmp=e,sum=1.0,p=1.0;
    int flag = -1;
    double i = 0;
    while (tmp >= e)
    {
        for (i = nn-1; i <= nn; i++)
        {
            p *= x;
            res *= i;
        }
        nn += 2;
        tmp = p / res;
        sum = sum + flag * tmp;
        flag = -flag;
    }
    return sum;
}

//6-9 使用函数统计指定数字的个数

#include 

int CountDigit(int number, int digit);

int main()
{
    int number, digit;

    scanf("%d %d", &number, &digit);
    printf("Number of digit %d in %d: %d\n", digit, number, CountDigit(number, digit));

    return 0;
}

int CountDigit(int number, int digit)
{
    int pows = 0, n = number;
    if (n < 0)
        n = -n;
    else if (n == 0 && digit == 0)
        return 1;
    while (n > 0)
    {
        if (n % 10 == digit)
            pows++;
        n /= 10;
    }
    return pows;
}

//6-10 空心的数字金字塔
 

#include 

void hollowPyramid(int n);

int main()
{
    int n;

    scanf("%d", &n);
    hollowPyramid(n);

    return 0;
}

void hollowPyramid(int n)
{
    for (int i = 0; i < n-1; i++)
    {
        for (int j = n-i-1; j > 0; j--)
        {
            printf(" ");
        }
        printf("%d", i+1);
        for (int k = 2 * i - 1; k >= 1; k--)
        {
            printf(" ");
        }
        if (i)
        {
            printf("%d", i+1);
        }
        printf("\n");
    }
    for (int i = 0; i < 2 * n - 1; i++)
    {
        printf("%d", n);
    }
}

//6-11 使用递归函数计算1到n之和
 

#include 

int sum(int n);

int main()
{
    int n;

    scanf("%d", &n);
    printf("%d\n", sum(n));

    return 0;
}

int sum(int n)
{
    if (n > 0)
        return n + sum(n - 1);
    else
        return 0;
}

//6-12 递归实现指数函数
 

#include 

double calc_pow(double x, int n);

int main()
{
    double x;
    int n;

    scanf("%lf %d", &x, &n);
    printf("%.0f\n", calc_pow(x, n));

    return 0;
}

double calc_pow(double x, int n)
{
    if (n > 1)
        return x * calc_pow(x, n - 1);
    else
        return x;
}

//6-13 递归计算Ackermenn函数
 

#include 

int Ack(int m, int n);

int main()
{
    int m, n;

    scanf("%d %d", &m, &n);
    printf("%d\n", Ack(m, n));

    return 0;
}

int Ack(int m, int n)
{
    if (!m)
        return n + 1;
    else if (n == 0 && m > 0)
        return Ack(m - 1, 1);
    else if (m > 0 && n > 0)
        return Ack(m - 1, Ack(m, n - 1));

}

//6-14 递归求Fabonacci数列
 

#include 

int f(int n);

int main()
{
    int n;

    scanf("%d", &n);
    printf("%d\n", f(n));

    return 0;
}

int f(int n)
{
    if (n == 0)
        return 0;
    if (n<=2&&n>0)
        return 1;
    else
        return f(n - 1) + f(n - 2);
}

//6 - 15 递归求简单交错幂级数的部分和

double fn(double x, int n);

int main()
{
    double x;
    int n;

    scanf("%lf %d", &x, &n);
    printf("%.2f\n", fn(x, n));

    return 0;
}

double fn(double x, int n)
{
    if (n == 1)
        return x;
    else
        return -x * fn(x, n - 1)+x;
}

//6-16 递归实现顺序输出整数

void printdigits(int n);

int main()
{
    int n;

    scanf("%d", &n);
    printdigits(n);

    return 0;
}

void printdigits(int n)
{
    int k = 1;
    if (n > 9)
    {
        k = n % 10;
        printdigits(n / 10);
        printf("%d\n", k);
    }
    else
        printf("%d\n", n);
}

//6-17 判断满足条件的三位数

#include 
#include 

int search(int n);

int main()
{
    int number;

    scanf("%d", &number);
    printf("count=%d\n", search(number));

    return 0;
}

1.普通方法

int search(int n)
{
	int i, cnt = 0;
	for (i = 101; i <= n; i++) {
		double digit = sqrt(i);
		int a = i / 100;
		int b = i / 10 % 10;
		int c = i % 10;
		if ((digit == (int)digit) && (a == b || a == c || b == c))
			cnt++;
	}
	return cnt;
}

 2.递归！！

int search(int n)
{
    if (n > 100 && (sqrt(n) ==(int) sqrt(n))&&(n/100==n%10||n/100==n/10%10||n%10==n/10%10))
        return 1 + search(n - 1);
    else if (n > 100)
        return 0 + search(n - 1);
    else
        return 0;
}
//错误点，else if中的定义错误了
//******************************
//** (n>100&&sqrt(n)!=(int)sqrt(n))
///开什么玩笑这是判断平方数的，不是题目中的返li

//6-18 递归求逆序数
 

#include 
# include 

unsigned int reverse(unsigned int number);

int main()
{
    unsigned int n;

    scanf("%u", &n);
    printf("%u\n", reverse(n));

    return 0;
}

失败品
//unsigned int reverse(unsigned int number)
//{
//    //static int k = 1 ;
//   
//    double p = 0.1;
//    while (m > 9)
//    {
//        k *= 10;
//        m /= 10;
//    }
//    if (number > 9)
//    {
//        p *= 10;
//        return number / k * (int)p + reverse(number-k*number);
//    }
//    else
//        return m;
//}






unsigned int reverse(unsigned int number)
{
    int po = 0, res = number;
    while (res) res /= 10, po++;

    if (number < 10) return number;
    return reverse(number / 10) + (number % 10) * pow(10, po - 1);
}

//6-19 十进制转换二进制
 

#include 

void dectobin(int n);

int main()
{
    int n;

    scanf("%d", &n);
    dectobin(n);

    return 0;
}

void dectobin(int n)
{
    if (n > 1)
    {
        dectobin(n / 2);
        printf("%d", n % 2);
    }
    else
        printf("%d", n);
}

//6-20 数组循环右移
 

#include 
#define MAXN 10

void ArrayShift(int a[], int n, int m);

int main()
{
    int a[MAXN], n, m;
    int i;

    scanf("%d %d", &n, &m);
    for (i = 0; i < n; i++) scanf("%d", &a[i]);

    ArrayShift(a, n, m);

    for (i = 0; i < n; i++) {
        if (i != 0) printf(" ");
        printf("%d", a[i]);
    }
    printf("\n");

    return 0;
}

void ArrayShift(int a[], int n, int m) {
    while (m--)
    {
        for (int i = n-1; i > 0 ; i--)
        {
            int tmp = a[i];
            a[i] = a[i -1];
            a[i -1] = tmp;
        }
    }
}

终于结束了，继续加油哦