簡單的指數題,從2開始,到sqrt(2147483648),之間每個數都要算。2算到31次方,而sqrt(2147483648)只用算到平方就可以了。
算幾次方很好確定,在紙上列個式子,那2來說,2^x=2147483648,那麼x=log(2147483648)/log(i),就可以辣``
還要注意去掉重複的,比如2^4=4^2=16
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
int zhishu[50000], k;
int i, j;
memset(zhishu,0,sizeof(zhishu));
k = 0;
for(i = 2 ; i <= 46342; i++)
{
for(j = 2 ; j <= log(2147483648)/log(i); j++)
{
zhishu[k++] = pow(i,j);
}
}
sort(zhishu,zhishu + k);
printf("%d\n",zhishu[1]);
for(i = 2 ; i < k ; i++)
{
if(zhishu[i] != zhishu[i-1])
{
printf("%d\n",zhishu[i]);
}
}
system("pause");
return 0;
}