常用算法及参考算法（1）累加（2）累乘（3）素数（4）最大公约数（5）最值问题（6）迭代法

常用算法及参考算法
（1）累加
（2）累乘
（3）素数
（4）最大公约数
（5）最值问题
（6）迭代法

在这里插入图片描述

1. 累加

#include <stdio.h>

int main() {
    int n, i, sum = 0;
    printf("Enter a positive integer n: ");
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
        sum += i;
    }
    printf("The sum from 1 to %d is: %d\n", n, sum);
    return 0;
}

2. 累乘

#include <stdio.h>

int main() {
    int n, i, product = 1;
    printf("Enter a positive integer n: ");
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
        product *= i;
    }
    printf("The product from 1 to %d is: %d\n", n, product);
    return 0;
}

3. 素数判断

#include <stdio.h>
#include <stdbool.h>

bool isPrime(int num) {
    if (num <= 1) return false;
    for (int i = 2; i * i <= num; i++) {
        if (num % i == 0) {
            return false;
        }
    }
    return true;
}

int main() {
    int num;
    printf("Enter a positive integer: ");
    scanf("%d", &num);
    if (isPrime(num)) {
        printf("%d is a prime number.\n", num);
    } else {
        printf("%d is not a prime number.\n", num);
    }
    return 0;
}

4. 最大公约数（最小公倍数可以通过两数乘积除以最大公约数得到）

#include <stdio.h>

int gcd(int a, int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

int main() {
    int a, b, g;
    printf("Enter two integers: ");
    scanf("%d %d", &a, &b);
    g = gcd(a, b);
    printf("The GCD of %d and %d is: %d\n", a, b, g);
    // LCM = (a * b) / GCD(a, b)
    printf("The LCM of %d and %d is: %d\n", a, b, (a * b) / g);
    return 0;
}

5. 最值问题（这里以找数组中的最大值为例）

#include <stdio.h>

int main() {
    int arr[] = {5, 2, 9, 1, 5, 6};
    int size = sizeof(arr) / sizeof(arr[0]);
    int max = arr[0];
    for (int i = 1; i < size; i++) {
        if (arr[i] > max) {
            max = arr[i];
        }
    }
    printf("The maximum value in the array is: %d\n", max);
    return 0;
}

6. 迭代法（这里以计算平方根为例，使用牛顿法）

#include <stdio.h>
#include <math.h>

double sqrt_iter(double x, double epsilon) {
    double guess = x / 2.0;
    while (fabs(guess * guess - x) >= epsilon) {
        guess = (guess + x / guess) / 2.0;
    }
    return guess;
}

int main() {
    double x, epsilon;
    printf("Enter a number and epsilon: ");
    scanf("%lf %lf", &x, &epsilon);
    printf("The square root of %.2lf is approximately %.5lf\n", x, sqrt_iter(x, epsilon));
    return 0;
}