一个面试题:
编写一个程序,该程序接收输入参数 'N'(一个无符号长整数),并打印两列输出。第一列以十六进制格式打印从1到N的数字;第二列打印左侧数字在二进制表示中出现的1的数量。条件是这个程序不应计算1的数量(因此不能对每个数字进行计算/没有除法运算符)。
我尝试通过利用0x0到0xF中的1的数量可以被重复使用来生成任何数字的1的数量。我将代码粘贴在下面(基本版,没有错误检查)。它可以给出正确的结果,但我对空间使用不满意。我该如何改进呢? (同时我不确定这是否是面试官想要的答案)。
void printRangeFasterWay(){
uint64_t num = ~0x0 ;
cout << " Enter upper number " ;
cin >> num ;
uint8_t arrayCount[] = { 0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4} ;
// This array will store information needed to print
uint8_t * newCount = new uint8_t[num] ;
uint64_t mask = 0x0 ;
memcpy(newCount, &arrayCount[0], 0x10) ;
uint64_t lower = 0;
uint64_t upper = 0xF;
uint64_t count = 0 ;
uint32_t zcount= 0 ;
do{
upper = std::min(upper, num) ;
for(count = lower ; count <= upper ; count++){
newCount[count] = (uint32_t)( newCount[count & mask] + newCount[(count & ~mask)>>(4*zcount)]) ;
}
lower += count ;
upper |= (upper<<4) ;
mask = ((mask<<4) | 0xF ) ;
zcount++ ;
}while(count<=num) ;
for(uint64_t xcount=0 ; xcount <= num ; xcount++){
cout << std::hex << " num = " << xcount << std::dec << " number of 1s = " << (uint32_t)newCount[xcount] << endl;
}
}
编辑添加样例运行
Enter upper number 18
num = 0 number of 1s = 0
num = 1 number of 1s = 1
num = 2 number of 1s = 1
num = 3 number of 1s = 2
num = 4 number of 1s = 1
num = 5 number of 1s = 2
num = 6 number of 1s = 2
num = 7 number of 1s = 3
num = 8 number of 1s = 1
num = 9 number of 1s = 2
num = a number of 1s = 2
num = b number of 1s = 3
num = c number of 1s = 2
num = d number of 1s = 3
num = e number of 1s = 3
num = f number of 1s = 4
num = 10 number of 1s = 1
num = 11 number of 1s = 2
num = 12 number of 1s = 2