题目1:
一个数如果恰好等于它的因子之和,这个数被成为”完数”,例如:6=1+2+3.请编程找出1000以内的完数

#include <stdio.h>

int main()
{
	int i,j;

	for(i=2;i<1000;i++)
	{
		int sum=0;

		for(j=1;j<i/2+1;j++)
		{
			if(i%j == 0)
			{
				sum += j;

			}
		}
			if(i == sum)
			{
				printf("%d\n",i);
			}
		}

		return 0;
}

题目2:
编写程序,打印出9×9乘法表

#include <stdio.h>

int main()
{
	int i,j;

	for(i=1;i<10;i++)
	{
		for(j=1;j<=i;j++)
		{
			printf("%d*%d=%d ",i,j,i*j);
		}
		printf("\n");
	}


	return 0;
}

题目3:
编写程序,打印菱形星号组合

#include <stdio.h>

int main()
{
	int x,m,y;

	for (y=10;y>=-10;y--)
	{
		m = y;
		for (x=0;x<=20;x++)
		{
			if (x == m || x == 20-m || x == -m || x == 20+m)
			{
				printf("*");
			}
			else
			{
				printf(" ");
			}
		}
		printf("\n");
	}

	return 0;
}

题目4:
输入一个字符串,计算字符串中子串出现的次字数

#include <stdio.h>

int count(char *p1,char *p2)
{
	int count=0;
	char *pa = p1;
	char *pb = p2;

	while(1)
	{
		if(*pa == 0)
		{
			break;
		}
		if(*pa == *pb)
		{
			pb++;
			if(*pb == 0)
			{
				count++;
				pb = p2;
			}
		}
		else
		{
			pb = p2;
		}
		pa++;
	}

	return count;
}

int main()
{
	char str[100] ={0};
	char str1[100]={0};

	printf("Input string:\n");//输入字符串
	scanf("%s",str);
	printf("Input string:\n");//输入子串
	scanf("%s",str1);

	printf("count=%d\n",count(str,str1));

	return 0;
}

题目5:
输入一个字符串,同时输入帧头和帧尾(可以是多个字符),将该字符串中合法的帧识别出来.
提示:帧头和帧尾分别是head和tail 字符串”asdheadhauboisoktail”中headhauboisoktail是合法帧

#include <stdio.h>

void fun(char *pstr)
{
	char headstr[10]="head";
	char tailstr[10]="tail";
	char *p  =pstr;
	char *ph =headstr;
	char *pt =tailstr;
	int  temp=1;


	while(temp == 1)
	{
		if(*p == 0)
		{
			temp = 0;
			break;
		}
		if(*p == *ph)
		{
			ph++;
			if(*ph == 0)
			{
				p=p-3;
				temp = 2;
				break;
			}
		}
		else
		{
			ph = headstr;
		}
		p++;
	}


	while(temp == 2)
	{
		if(*p == 0)
		{
			temp = 0;
			break;
		}

		*pstr++=*p;

		if(*p == *pt)
		{
			pt++;
			if(*pt == 0)
			{
				*(++p)='\0';
				*pstr='\0';
				temp=3;
				break;
			}
		}
		else
		{
			pt = tailstr;
		}
		p++;
    	}
}

int main()
{
	char str[100]={0};
	printf("Input string:\n");
	scanf("%s",str);
	fun(str);
	printf("%s\n",str);

	return 0;
}

题目6
有一分数序列:2/1,3/2,5/3,8/5,13/8,21/13…求出这个数列的前20项之和。

#include <stdio.h>

int main()
{
	int n;
	float a=2,b=1;
	float sum=0;

	for(n=0;n<20;n++)
	{
		sum+=a/b;
		a=a+b;
		b=a-b;
	}

	printf("%f\n",sum);

	return 0;
}

题目7
打印出所有的“水仙花数”,所谓“水仙花数”是指一个三位数,其各位数字立方和等于该数本身。例如:153是一个“水仙花数”,因为153=1的三次方+5的三次方+3的三次方。

#include <stdio.h>

int main()
{
	int 1;
	int j1,j2,j3;

	for(i=100;i<=999;i++)
	{
		j1=i%10;
		j2=(i/10)%10;
		j3=i/100;
		if(j1*j1*j1+j2*j2*j2+j3*j3*j3 == i)
			printf("%d\n",i);
	}

    	return 0;
    }

题目8
求s=a+aa+aaa+aaaa+aa…a的值,其中a是一个数字。例如2+22+222+2222+22222(此时共有5个数相加),几个数相加由键盘控制

#include <stdio.h>

int main()
{
	int a,n,sum=0;
	int i;

	printf("Input a  n\n");
	scanf("%d,%d",&a,&n);

	for(i=1;i<=n;i++)
	{
		sum=sum*10+i*a;
	}

	printf("sum = %d\n",sum);

	return 0;
}

题目9
将字符串“We Are Family!”,去除空格后打印出来。(WeAreFamily!)

#include <stdio.h>

void fun(char *pstr)
{
	char *p=pstr;
	int   temp=1;

	while(1)
	{
		if(*p == '\0')
		{
			*pstr='\0';
			break;
		}
		if(*p == ' ')
		{
			p++;
			continue;
		}
		*pstr++=*p++;
	}

}

int main()
{
	char str[100]="We Are Family!";

	printf("%s\n",str);

	fun(str);

	printf("%s\n",str);

	return 0;
}

题目10
输入一个字符串,判断其是否是回文。(回文:即正读和反读都一样,如abcba, abccba)

#include <stdio.h>

int fun(char *pstr)
{
	int left  = 0;
	int len=sizeof(pstr)/sizeof(int);
	int right = len-1;

	while(left < right)
	{
		if(pstr[left] != pstr[right])
		{
			return 0;
		}
		left++;
		right--;
	}

	return 1;

}

int main()
{
	char str[100]="aaabbbbbbbccccbbbbbbbaaa";
	int temp=fun(str);

	if(temp == 1)
	{
		printf("%s 是回文\n",str);
	}
	else
	{
		printf("%s 不是回文\n",str);
	}


	return 0;
}

题目12
输入一段字符串,无论是否有重复字母出现,都只打印出现过的小写字母,并按照小写字母顺序打印。
(如输入qewqwr322rqw<>211qESFSSEraZz, 打印aeqrwz)

#include <stdio.h>

int main()
    {
    	char str[256] = {0};
    	char c;
    	char i;


	while(c != '\n')
	{
		scanf ("%c", &c);
		str[c]++;
	}

	for(i='a';i<'z';i++)
	{
		if(str[i] != 0 || str[i+'A'-'a'] != 0)
		{
			printf("%c",i);
		}
	}
	printf("\n");

	return 0;
}
11-30 15:03