个人贡献

一、简述你完成的工作
1.协助完成数据库逻辑结构的设计，并撰写了部分数据库的代码
2.部分前端代码的修改与补充
3.系统测试与debug

二、你贡献的代码行数？相关代码链接？
我贡献的代码行数：1004行
代码链接：
admin部分
addDao.jsp: https://gitee.com/zhang-yibo/micro-certification/blob/master/fileweb/web/files/admin/addDao.jsp
delDao.jsp: https://gitee.com/zhang-yibo/micro-certification/blob/master/fileweb/web/files/admin/delDao.jsp
update.jsp: https://gitee.com/zhang-yibo/micro-certification/blob/master/fileweb/web/files/admin/update.jsp
updateDao.jsp: https://gitee.com/zhang-yibo/micro-certification/blob/master/fileweb/web/files/admin/updateDao.jsp
file部分
tlist.jsp: https://gitee.com/zhang-yibo/micro-certification/blob/master/fileweb/web/files/file/tlist.jsp
upload.jsp: https://gitee.com/zhang-yibo/micro-certification/blob/master/fileweb/web/files/file/upload.jsp

三、你们小组总共的文档数？你贡献的文档数？相关链接？
小组总文档数：7
我贡献的文档数：4（共同完成）
团队作业一 —— 小组成员介绍：https://www.cnblogs.com/sanfeng-ooo/p/16838964.html

 
#include <stdio.h>
#include <openssl/bn.h>

int main() {
    BIGNUM *result = BN_new();
    BN_CTX *ctx = BN_CTX_new();
    BN_one(result);
    int i, j, prime;
    for (i = 2; i <= 1000; i++) {
        prime = 1;
        for (j = 2; j < i; j++) {
            if (i % j == 0) {
                prime = 0;
                break;
            }
        }
        if (prime) {
            BIGNUM *bn_i = BN_new();
            BN_set_word(bn_i, i);
            BN_mul(result, result, bn_i, ctx);
            BN_free(bn_i);
        }
    }
    char *result_str = BN_bn2dec(result);
    printf("The product of primes under 1000 is: %s\n", result_str);
    OPENSSL_free(result_str);
    BN_free(result);
    BN_CTX_free(ctx);
    return 0;
}

#include <stdio.h>
#include <openssl/bn.h>

int main()
{
    BIGNUM *result = BN_new();
    BIGNUM *temp = BN_new();
    BN_CTX *ctx = BN_CTX_new();

    BN_set_word(result, 1);

    char *nums[] = {"20201319", "20201320", "20201321", "20201322", "20201323", "20201324", "20201325", "20201326", "20201327", "20201328", "20201329", "20201330", "20201331", "20201332", "20201301", "20201302"};

    for (int i = 0; i < 16; i++)
    {
        BN_dec2bn(&temp, nums[i]);
        BN_mul(result, result, temp, ctx);
    }

    char *result_str = BN_bn2dec(result);
    printf("%s\n", result_str);

    BN_free(result);
    BN_free(temp);
    BN_CTX_free(ctx);

    return 0;
}

《需求规格说明书》：https://www.cnblogs.com/sanfeng-ooo/p/16841766.html
团队作业二 —— 需求分析：https://www.cnblogs.com/sanfeng-ooo/p/16841842.html
团队作业四 —— 描述设计：https://www.cnblogs.com/sanfeng-ooo/p/16907872.html
团队作业 —— 贡献排序：https://www.cnblogs.com/sanfeng-ooo/p/16964655.html

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

int main() {
    mpz_t product, num;
    mpz_init(product);
    mpz_init(num);
    int i, j, is_prime;

    for (i = 2; i <= 10000; i++) {
        is_prime = 1;
        for (j = 2; j <= i / 2; j++) {
            if (i % j == 0) {
                is_prime = 0;
                break;
            }
        }
        if (is_prime) {
            mpz_set_ui(num, i);
            mpz_mul(product, product, num);
        }
    }

    gmp_printf("The product of primes up to 10000 is %Zd\n", product);

    mpz_clear(product);
    mpz_clear(num);
    return 0;
}

https://www.cnblogs.com/51ggh/p/16359348.html