Question

c语言编程：求1000--9999之间所有回文数的个数（如1111、1221、4994等均为回文数）

Answer 1

#include <iostream>
#include <string>
using namespace std;
void main()
{
for(unsigned int i = 1000; i <坦告= 9999; i++)
{
char zzz[100];
sprintf(zzz,"%i"让带明,i);
string lol = zzz;
if(lol[0] == lol[3] && lol[1] == lol[2])
{
cout <行肢< lol << endl;
}
}
}

Answer 2

存成字符型直接一个一个的判断
还不用区分下奇数和偶数情况呢

Answer 3

真毛

Answer 4

//---------------------------------------------------------------------------

#include <盯正stdio.h>

int ch(int b)
{
char a[5];
sprintf(a,"%d",b);
if (a[0]==a[3]&&a[1]==a[2])
return 1;

return 0;
}
int main(void)
{
int i;
for (i=1000; i<罩则宏10000; i++)
if (ch(i)) printf("%d\物册n",i);
return 0;
}
//---------------------------------------------------------------------------