用C语言编写其中的题目

好的，以下是每个实验题目的C语言实现：

### 题目1：生成并显示一个大写字母，并用其显示特定图案

**项目名**: ex55
**源文件名**: exc55.c

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main() {
    char ch;
    srand(time(NULL));
    ch = 'A' + rand() % 26;

    printf("Generated Character: %c\n", ch);

    for (int i = 1; i <= 5; i++) {
        for (int j = 1; j <= i; j++) {
            printf("%c ", ch);
        }
        printf("\n");
    }

    return 0;
}

### 题目2：找出并显示1000之内的所有完数

**项目名**: ex56
**源文件名**: exc56.c

#include <stdio.h>

int isPerfectNumber(int num) {
    int sum = 0;
    for (int i = 1; i <= num / 2; i++) {
        if (num % i == 0) {
            sum += i;
        }
    }
    return sum == num;
}

int main() {
    printf("Perfect numbers within 1000:\n");
    for (int i = 1; i <= 1000; i++) {
        if (isPerfectNumber(i)) {
            printf("%d\n", i);
        }
    }
    return 0;
}

### 题目3：计算分数序列的前20项之和

**项目名**: ex57
**源文件名**: exc57.c

#include <stdio.h>

double fractionSum(int n) {
    double sum = 0.0;
    int numerator = 1, denominator = 2;
    for (int i = 0; i < n; i++) {
        sum += (double)numerator / denominator;
        numerator += 1;
        denominator += 2;
    }
    return sum;
}

int main() {
    double sum = fractionSum(20);
    printf("The sum of the first 20 terms of the sequence is: %.10f\n", sum);
    return 0;
}

### 题目4：计算Sn = a + aa + aaa + ... + aa...a

**项目名**: ex58
**源文件名**: exc58.c

#include <stdio.h>

long long computeSn(int a, int n) {
    long long sum = 0, term = 0;
    for (int i = 0; i < n; i++) {
        term = term * 10 + a;
        sum += term;
    }
    return sum;
}

int main() {
    int a, n;
    printf("Enter the values of a and n: ");
    scanf("%d %d", &a, &n);

    long long result = computeSn(a, n);
    printf("The value of S%d is: %lld\n", n, result);
    return 0;
}

### 题目5：求最大公约数和最小公倍数（选做）

**项目名**: ex59
**源文件名**: exc59.c

#include <stdio.h>

int gcd(int a, int b) {
    while (b != 0) {
        int temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

int lcm(int a, int b) {
    return (a / gcd(a, b)) * b;
}

int main() {
    int m, n;
    printf("Enter two positive integers: ");
    scanf("%d %d", &m, &n);

    int g = gcd(m, n);
    int l = lcm(m, n);

    printf("GCD of %d and %d is: %d\n", m, n, g);
    printf("LCM of %d and %d is: %d\n", m, n, l);
    return 0;
}

### 题目6：求满足abcd + cadb = 7856的所有四位数对（选做）

**项目名**: ex510
**源文件名**: exc510.c

#include <stdio.h>

void findPairs() {
    for (int abcd = 1000; abcd <= 9999; abcd++) {
        int a = abcd / 1000;
        int b = (abcd / 100) % 10;
        int c = (abcd / 10) % 10;
        int d = abcd % 10;
        int cadb = c * 1000 + a * 100 + d * 10 + b;

        if (abcd + cadb == 7856) {
            printf("Pair found: %d and %d\n", abcd, cadb);
        }
    }
}

int main() {
    printf("Finding pairs of four-digit numbers such that abcd + cadb = 7856:\n");
    findPairs();
    return 0;
}

希望这些代码对你有帮助！如果有任何问题或需要进一步解释，请告诉我。