求 1到100000000 既是回文又是素数的数字

# include<stdio.h>
# include<stdlib.h>
# include<string.h>
# include<math.h>
unsigned long data[2000] ;
int count;
char str[20];
char str1[20];
char str2[20];

int cmp(const void *a,const void *b)
{
return *(unsigned long *)a-*(unsigned long *)b;
}

int issushu(unsigned long num)
{

unsigned long i,j ,k = num;

for (j = (unsigned long)sqrt(k),i = 2;i <= j+1;i++)
if (num%i==0)
return 0;

return 1;

}
void _itoah(int n,int digit )
{
int i = 0;
for(i = 0;i < digit;i ++)
{
//str[digit-i-1] =  n%10 + '0';
str[digit-i-1] =  n%10 + '0';
n /=10;
}

}
void _itoad(int n,int digit,int b )// 针对 奇数和偶数的不同 增加一个标志位
{
int i = 0,j = digit;
if(b==1)
digit ++;
for(i = 0;i < j;i ++)
{
str[digit++] =  n%10 + '0';
n /=10;
}
}
void even_huiwen(int n)
{
long i ,j,digit = n/2,tmp = 0;
//j = (long)pow(10,digit) -1;
j = (long)pow(10,digit) -1;
for(i = (long)pow(10,digit-1); i <= j;i++ ) //i代表偶数的高位//(long)pow(10,n-1); i <= j;i++ )
{

_itoah(i,digit);
_itoad(i,digit,0);
str
= '/0';
strcpy(str1,str);
strrev(str1);
if (!strcmp(str1,str))  if(issushu(atol(str)))
data[count ++] = atol(str);//printf("%ld/t",atol(str)),data[count ++] = atol(str);

memset(str,NULL,sizeof(str));
memset(str1,NULL,sizeof(str1));

}
}
void odd_huiwen(int n)
{
long i ,j,k,digit = n/2,tmp = 0;
j = (long)pow(10,digit) -1;

for(k = 0;k < 10; k ++) // 奇数位中间的数字
for (i = (long)pow(10,digit-1);i <= j ; i ++)
{
_itoah(i,digit);
str[digit] = k + '0';
_itoad(i,digit,1);
str
= '/0';
strcpy(str1,str);
strrev(str1);
if (!strcmp(str1,str)) if(issushu(atol(str)))
data[count ++] = atol(str);//printf("%ld/t",atol(str)),data[count ++] = atol(str);

memset(str,NULL,sizeof(str));
memset(str1,NULL,sizeof(str1));
}
}
int  main()
{
unsigned long a,i,j,x,y;
data[0] = 2,data[1] = 3,data[2] = 5,data[3] = 7,data[4] = 11;
//freopen("pprime1.out","w",stdout);
count = 5;
int n;

//生产所有的回文素数
for(int _digit = 3; _digit <= 8;_digit+=2)
{
if(_digit%2==0)
even_huiwen(_digit);
else
odd_huiwen(_digit);
}

qsort(data,count,sizeof(unsigned long ),cmp);
scanf("%d",&n);
while (n--)
{
scanf("%ld%ld",&x,&y);
for (i=0;data[i] <=y&&i<=count;i++)
{
if (data[i]>=x)
printf("%ld/n",data[i]);
}

}

return 0;

}