Annie's Homework

level1/p03_Diophantus/Diophantus.c

@@ -0,0 +1,31 @@
+#include<stdio.h>
+
+int main()
+{
+        int Diophantus_age;
+        int childhood, youth, bachelor, before_having_son, between_death, son_age;
+        int numerator, denominator;
+
+        //人生中某些阶段占总年龄比例的倒数
+        childhood = 6;
+        youth = 12;
+        bachelor = 7;
+        son_age = 2;
+
+        //人生中某些阶段经历的年数
+        before_having_son = 5;
+        between_death = 4;
+
+        //解方程
+        numerator = (before_having_son + between_death) * childhood * youth * bachelor * son_age;
+        denominator = childhood * youth * bachelor * son_age
+                                - childhood * youth * bachelor
+                                - childhood * youth * son_age
+                                - childhood * bachelor * son_age
+                                - youth * bachelor * son_age;
+        Diophantus_age = numerator / denominator;
+
+        printf("%d\n", Diophantus_age - between_death);
+
+        return 0;
+}

level1/p04_ narcissus/narcissus.c

@@ -0,0 +1,22 @@
+#include<stdio.h>
+
+int main()
+{
+        int a, b, c;    //个位数、十位数、百位数
+        int num = 100;        //当前枚举的数
+
+        while(num >= 100 && num <= 999)
+        {
+                c = num / 100;
+                b = num / 10 % 10;
+                a = num % 10;
+
+                if(num == a * a * a + b * b * b + c * c * c)
+                        printf("%d ", num);
+
+                ++num;
+        }
+        printf("\n");
+
+        return 0;
+}

level1/p05_allPrimes/all_primes.c

@@ -0,0 +1,55 @@
+//打印2~1000以内的所有素数
+//打印出总的计算时间
+//尽可能地优化算法的效率
+
+#include<stdio.h>
+#include<stdlib.h>
+#include<time.h>
+
+int main()
+{
+        int n = 2;              //当前检索的数
+        int prime[200] = {0};   //用于存放找到的素数
+        int i;
+
+        clock_t start, finish;
+        double duration;
+        //测量一个事件的持续时间
+
+        start = clock();        //计时开始
+
+        prime[0] = 2;
+
+        //若已检索出的任何一个素数都不是当前检索的数的因子
+        //那么当前检索的数是素数
+        while(++n <= 1000)
+        {
+                for(i = 0; prime[i]; ++i)
+                {
+                        if(n % prime[i] == 0)
+                                break;
+                }
+
+                if(prime[i] == 0)
+                        prime[i] = n;
+        }
+
+        finish = clock();       //计时结束
+        duration = (double)(finish - start) / CLOCKS_PER_SEC;
+
+        //打印检索到的素数
+        i = -1;
+        while(prime[++i])
+        {
+                printf("%5d", prime[i]);
+
+                //每15个素数输出为一行
+                if((i + 1) % 15 == 0)
+                        printf("\n");
+        }
+
+        //打印总的计算时间
+        printf("\n\nTime to calculate all primes is %f seconds.\n", duration);
+
+        return 0;
+}

level1/p06_Goldbach/Goldbach.c

@@ -0,0 +1,68 @@
+//在100范围内验证哥德巴赫猜想的正确性
+//原初猜想的现代陈述：任一大于5的整数都可写成三个质数之和
+
+#include<stdio.h>
+#define N 100         //上界
+#define M 6           //下界
+
+int main()
+{
+        int i, j1, j2, j3;
+        int prime[50] = {0};            //用于存放100以内的质数
+        void Prime(int prime[]);        //把100以内的素数存放到prime数组里
+
+        Prime(prime);
+
+        for(i = M; i <= N; ++i)
+        {
+                //如果验证了当前数符合哥德巴赫猜想，就跳到最外层循环的末尾
+                for(j1 = 0; prime[j1] <= (i - 4); ++j1)
+                        for(j2 = j1; prime[j2] <= (i - 4); ++j2)
+                                for(j3 = j2; prime[j3] <= (i - 4); ++j3)
+                                        if(i == prime[j1] + prime[j2] + prime[j3])
+                                                goto sf;
+
+                //如果没有跳到末尾，则当前数不符合哥德巴赫猜想，输出"Error!"，终止程序并返回“1”
+                printf("Error!\n");
+                return 1;
+
+                sf:
+                        ;
+        }
+
+        printf("Goldbach conjecture within 100 is verified.\n");
+        return 0;
+}
+
+void Prime(int prime[])
+{
+        int i, j;
+        int num;                        //当前检索的数
+
+        prime[0] = 2;
+        i = 0;
+        num = 2;
+
+        while(++num <= N)
+        {
+                for(j = 0; prime[j]; ++j)
+                {
+                        if(num % prime[j] == 0)
+                                break;
+                }
+
+                if(prime[j] == 0)
+                        prime[j] = num;
+        }
+
+        //这段用来测试，完成后删去
+        j = -1;
+        while(prime[++j])
+        {
+                printf("%5d ", prime[j]);
+
+                if((j + 1) % 10 == 0)
+                        printf("\n");
+        }
+        printf("\n\n");
+}

level1/p07_encrypt_decrypt/decrypt.c

@@ -0,0 +1,156 @@
+#include<stdio.h>
+#include<string.h>
+#include<stdlib.h>
+
+typedef unsigned long UL;
+
+extern const int LEN2;
+
+UL n, d;
+
+void decrypt(char cipher[LEN2])
+{
+        UL decrypt_single(UL c);
+
+        //获取私钥
+        FILE *fp = fopen("private_key.txt", "r");
+        if(fp == NULL) {printf("Error004!\n"); exit(EOF);}
+        fscanf(fp, "n = %lu", &n);
+        while(fgetc(fp) != '\n') ;
+        fscanf(fp, "d = %lu", &d);
+        while(fgetc(fp) != '\n') ;
+        fclose(fp);
+
+        //test
+        printf("n = %lu\td = %lu\n", n, d);
+
+        //n的长度
+        int len_n = 0;
+        UL temp_n = n;
+        while(temp_n)
+        {
+                temp_n = temp_n / 10;
+                ++len_n;
+        }
+
+        //test
+        printf("len_n = %d\n", len_n);
+
+        //密文的长度
+        const int len_cip = strlen(cipher);
+
+        //把密文还原成一段只包含数字的字符串
+        char num_str[LEN2];
+        for(int i = 0; i < LEN2; ++i)
+        {
+                num_str[i] = '\0';
+        }
+        int ns_i = 0;   //num_str数组的下标
+        int temp_i;     //把字母还原为数值
+        for(int i = 0; i < len_cip; ++i)
+        {
+                if(cipher[i] >= 'a' && cipher[i] <= 'z')
+                {
+                        temp_i = cipher[i] - 97;
+                        num_str[ns_i] = temp_i / 10;
+                        num_str[ns_i + 1] = temp_i % 10;
+                        ns_i = ns_i + 2;
+                }
+                else if(cipher[i] >= 'A' && cipher[i] <= 'Z')
+                {
+                        temp_i = cipher[i] - 39;
+                        num_str[ns_i] = temp_i / 10;
+                        num_str[ns_i + 1] = temp_i % 10;
+                        ns_i = ns_i + 2;
+                }
+                else
+                {
+                        num_str[ns_i] = cipher[i] - '0';
+                        num_str[ns_i + 1] = cipher[i + 1] - '0';
+                        ns_i = ns_i + 2;
+                        ++i;
+                }
+        }
+
+        //test
+        int j_j = 0;
+        printf("num_str : ");
+        for(int i = 0; i < 5; ++i)
+        {
+                int temp_j = j_j + len_n;
+                for(; j_j < temp_j; ++j_j)
+                {
+                        printf("%d", (int)num_str[j_j]);
+                }
+                printf("\t");
+        }
+        printf("\n");
+
+        //明文的长度
+        int len = ns_i / len_n;
+
+        //test
+        printf("len = %d\n", len);
+
+        //把上一步得到的字符串分割成单个的数
+        UL num[len];
+        UL temp_num;
+        UL t10;
+        for(int i = 0; i < len; ++i)
+        {
+                temp_num = 0;
+                t10 = 1;
+                for(int j = (i + 1) * len_n - 1; j >= i * len_n; --j)
+                {
+                        temp_num = temp_num + (int)num_str[j] * t10;
+                        t10 = t10 * 10;
+                }
+                num[i] = temp_num;
+        }
+
+        //test
+        printf("num : ");
+        for(int i = 0; i < len; ++i)
+        {
+                printf("%lu ", num[i]);
+        }
+        printf("\n");
+
+        //还原字符串
+        char plain[len + 1];
+        plain[len] = '\0';
+        printf("temp_result : ");//test
+        for(int i = 0; i < len; ++i)
+        {
+                //test
+                UL temp_result = decrypt_single(num[i]);
+                printf("%lu ", temp_result);
+
+                plain[i] = (char)temp_result;
+        }
+        printf("\n");//test
+
+        //test
+        printf("The ASCII of plain : ");
+        for(int i = 0; i < len; ++i)
+        {
+                printf("%d ", (int)plain[i]);
+        }
+        printf("\n");
+
+        //输出
+        printf("The plain text is: \n");
+        printf("%s\n\n", plain);
+}
+
+UL decrypt_single(UL c)
+{
+        UL result = 1;
+        for(UL i = 0; i < d; ++i)
+        {
+                result = result * c;
+                result = result % n;
+        }
+
+        return result;
+}