Homework 1-6 by LuXiaoya

README.md

@@ -78,3 +78,7 @@
 ### 其他
 
 [免费的计算机编程类中文书籍](https://github.com/wwj718/free-programming-books-zh_CN)
+
+github打不开
+
+打开CMD运行如下命令 ipconfig /flushdns

level1/p01_runningLetter/level1.1.cpp

@@ -0,0 +1,48 @@
+#include <stdio.h>
+#include <windows.h>
+#include <stdlib.h>
+#include <string.h>
+
+int main()
+{
+	system("mode con cols=50 lines=25 ");
+	char a[100]={0};
+	int i,j,p;
+	gets(a);
+	for(i=0;i<50-strlen(a);i++)
+	{
+		for(j=0;j<i;j++)
+		{
+			printf(" ");	
+		};
+		puts(a);
+		Sleep(200);
+		system("cls");						
+	}
+	for(int p=i;p>0;p--)
+	{
+		for(int j=0;j<p;j++)
+		{
+			printf(" ");	
+		};
+		puts(a);
+		Sleep(200);
+		system("cls");						
+	}
+
+
+
+}
+//#include<stdio.h>
+//#include<sys/types.h>
+//#include<sys/ioctl.h>
+//#include<unistd.h>
+//#include<termios.h>
+//int main()
+//{
+//struct winsize size;
+//ioctl(STDIN_FILENO,TIOCGWINSZ,&size);
+//printf("%d\n",size.ws_col);
+//printf("%d\n",size.ws_row);
+//return 0;
+//}

level1/p02_isPrime/level1.2.c

@@ -0,0 +1,15 @@
+#include <stdio.h>
+int main()
+{
+	int n, i;
+	scanf_s("%d", &n);
+	for (i = 2;i < n;i++)
+	{
+		if (0==n%i)break;
+	}
+	if (i == n)
+		printf(" %d 是素数", n);
+	else
+		printf(" %d 不是素数", n);
+	return 0;
+}

level1/p03_Diophantus/README.md

@@ -14,4 +14,8 @@
 
 年级是他的一半。
 
-问儿子死时丢番图多大？
+问儿子死时丢番图多大？
+
+
+
+算法问题：怎么解方程，年龄是整数还要考虑相除后取整带来的误差

level1/p03_Diophantus/level1.3.cpp

@@ -0,0 +1,16 @@
+#include <stdio.h>
+
+int main()
+{
+	int m=0,i=0,k;
+	do
+	{
+		i++;
+		if(i%12==0&&i%7==0&&(i/6+i/12+i/7+5+i/2+4)==i)//ֱ����ѭ���ⷽ�̳���ȡ�����õ�����������Ҫ��֤������6,12,7�Ĺ�����
+		{
+			m=i;
+		}
+	}
+	while(m!=i);
+	printf("������ʱ����ͼ��������%d",m-4);
+} 

level1/p04_ narcissus/level1.4.cpp

@@ -0,0 +1,24 @@
+#include <stdio.h>
+#include <math.h>
+int main()
+{
+	int m,s,t=0,k,i;
+	printf("��λ��ˮ�ɻ����У�"); 
+	for(m=100;m<1000;m++)
+	{
+	    k=m;
+		s=0;//s��i��k��Ҫ��ʼ����ע���ʼ����λ��Ӧ����ѭ���ڻ�����	
+		for(i=0;k>0;i++)
+		{ 
+			t=k%10;
+			s+=pow(t,3);
+			k=k/10;	
+		}
+		if(m==s)
+		{
+			printf("%d\t",m);
+		}
+	}
+
+	return 0;
+} 

level1/p05_allPrimes/level1.5.c

@@ -0,0 +1,89 @@
+/*节省时间:
+- break；
+- 只用奇数，排除偶数,n+=2
+- i<sqrt(n)就可以结束了
+- 用数组记录质数，然后直接循环除质数就可以了，不用把所有的数循环一遍
+*/
+
+#include <stdio.h>
+#include <time.h>
+#include <math.h>
+
+void isprime_1();
+void isprime_2();
+
+int main()
+{
+	clock_t start_t, finish_t;
+	double total_t;
+
+	start_t = clock();
+	isprime_1();
+	finish_t = clock();
+	total_t = (double)(finish_t - start_t) / CLOCKS_PER_SEC;
+	printf("\n质数数组法运行总时间是%f 秒\n", total_t);
+
+	start_t = clock();
+	isprime_2();
+	finish_t = clock();
+	total_t = (double)(finish_t - start_t) / CLOCKS_PER_SEC;
+	printf("\n直接循环取余程序运行总时间是%f 秒\n", total_t);
+	return 0;
+}
+
+void isprime_1()
+{
+	int x = 0, isprime, n = 3, a[1000] = { 2,0 };
+
+	while (n < 1000)
+	{
+		isprime = 1;
+		for (int i = 0;a[i] < sqrt(n);i++)
+		{
+			if (n % a[i]==0)
+			{
+				isprime = 0;
+				break;
+			}
+		}
+		if (isprime)
+		{
+			a[x + 1] = n;
+			x += 1;
+		}
+		n += 2;
+	}
+
+
+	for (int i = 0;i <= x;i++)
+	{
+		printf("%d\t", a[i]);
+		if ((i + 1) % 10 == 0)
+		{
+			printf("\n");
+		}
+	}
+	printf("\n");
+
+}
+
+void isprime_2()
+{
+	int n = 0, i, j, isprime;
+	for (i = 2;i <= 1000; i++)
+	{
+		isprime = 1;
+		for (j = 2; j <= sqrt(i); j++)
+			if (i % j == 0)
+			{
+				isprime = 0;
+				break;
+			}
+		if (isprime)
+		{
+			n++;
+			if (n % 10 == 0) printf("%d\n", i);
+			else printf("%d\t", i);
+		}
+	}
+}

level1/p06_Goldbach/level1.6.cpp

@@ -0,0 +1,70 @@
+//��°ͺղ��룺��һ����2��ż������д����������֮��
+#include <stdio.h>
+#include <math.h>
+
+void prime(int a[]);//����һ�� ������100���ڵ����� 
+int isprime(int q,int a[]);//���������ȶ��ж��Dz�������
+
+//n��ȥһ�������ж��Ƿ��в�ֵΪ���������У����֤ 
+int main()
+{
+	int t,n=4;
+	int p[100]={0};
+	prime(p);
+	while(n<=100)
+	{
+		for(int i=0;p[i]>0;i++)
+		{
+			if(isprime(n-p[i],p))
+			{
+				printf("%d=%d+%d\n",n,p[i],n-p[i]);
+				break;
+			}
+			if(p[i]==0)
+			{
+				printf("Error");
+			} 
+		}
+		n+=2;		
+	}
+	printf("If there is no error,Goldbach conjecture has been proved.")
+	return 0;	
+
+}
+
+void prime(int a[])
+{
+	int x=0,isprime,n=3;
+	while(n<100)
+	{
+		isprime=1;
+		a[0]=2;
+	    for(int i=0;a[i]<sqrt(n);i++)
+	    {
+		    if(n%a[i]==0)
+	    	{
+		    	isprime=0;
+		    	break;
+		    }
+	    }
+	    if(isprime)
+	    {
+		    a[x+1]=n;
+		    x+=1;
+	    }	
+		n+=2;		
+	}
+}
+
+int isprime(int q,int a[])
+{
+	int isprime=0;
+    for(int i=0;a[i]>0;i++)
+    {
+    	if(q==a[i])
+    	{
+    		isprime=1;
+		} 
+    }
+    return isprime;
+}