[LeetCode 231 326 342] Power of Number 2,3,4

//

#include "stdafx.h"
#include <stdio.h>
#include <math.h>

int isPowerOfTwo(int n)//231
{
    if(n<=0)return 0;
    if(0==(n&(n-1))){
        return 1;
    }else
        return 0;
}

int isPowerOfThree(int n)//326
{ //int范围内最大的3的幂是多少？1162261467
    //3 9 27 81
    //big3 % n ==0
    //big3=3^k
    //k=log3(maxint)
    if(n<=0)return 0;
    const int maxint=0x7FFFFFFF;
    int k=(int)log((long double)maxint)/log((long double)3);//这种方法仅对素数有用
    int big3=pow((double)3,(double)k);
    printf("big3 is %d\n",big3);
    return (big3%n ==0);
}

int isPowerOfFour(int n)//342
{
    //2 8 false 2=10  8=1000
    //4 16 true
    //5=101

    //0101 & 8 1000 = 0000
    //0101 & 4 0100 = 0100
    if(n<=0)return 0;
    if((n&(n-1))==0 && (n&0x55555555)){
        return 1;
    }else{
        return 0;
    }
}

int _tmain(int argc, _TCHAR* argv[])
{
    int x;
    scanf("%d",&x);
    do{
        if(isPowerOfTwo(x)){
            printf("x is power of 2\n");
        }else{
            printf("x is not power of 2\n");
        }
        if(isPowerOfThree(x)){
            printf("x is power of 3\n");
        }else{
            printf("x is not power of 3\n");
        }
        if(isPowerOfFour(x)){
            printf("x is power of 4\n");
        }else{
            printf("x is not power of 4\n");
        }
        scanf("%d",&x);
    }while(x!=-1);
    return 0;
}