我的函数 isOre 正在检查数字是否为“谐波除数(矿石数)”(https://en.wikipedia.org/wiki/Harmonic_divisor_number#CITEREFCohenSorli2010)。我无法正确比较“结果变量”是否为整数,因为它的一部分在指数内。(https://www.h-schmidt.net/FloatConverter/IEEE754.html)。
我的问题是:1.我该如何解决?2. 是否有替代解决方案?
#include <iostream>
using namespace std;
bool isOre(unsigned int n){
double numbersSum = 0;
int number = 0;
for (int i = 1; i <= n; ++i){
if (n % i == 0){
numbersSum += (double)1 / i ;
//cout << "numbersSum " << numbersSum << endl;
++number;
//cout << "number " << number << endl;
}
}
double result = number / numbersSum;
int result1 = result;
cout << " " << result << " " << result1 << " ";
if (result == result1) return true;
else return false;
}
int main() {
//Test sequence of Ore numbers from https://oeis.org/A001599
int array[34] = {1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128,
8190, 18600, 18620, 27846, 30240, 32760, 55860, 105664, 117800, 167400,
173600, 237510, 242060, 332640, 360360, 539400, 695520, 726180, 753480, 950976,
1089270, 1421280, 1539720};
cout << "array[i]" << " " << "double" << " " << "toInt" << " " << "Result" << endl;
for (int i = 0; i < 34; ++i){
cout << array[i];
cout << isOre(array[i]) << endl;
}
return 0;
}
主要是我的测试用例序列必须通过。
我的输出:
array[i] double toInt 结果
1 1 1 1
6 2 2 1
28 3 3 1
140 5 5 1
270 6 6 1
496 5 5 1
672 8 8 0
1638 9 9 0
2970 11 11 0
6200 10 10 1
8128 7 7 1
8190 15 15 0
18600 15 15 0
18620 14 14 0
27846 17 17 0
30240 24 23 0
32760 24 24 0
55860 21 21 0
105664 13 13 0
117800 19 19 0
167400 27 27 0
173600 25 25 0
237510 29 29 0
242060 26 26 0
332640 44 43 0
360360 44 43 0
539400 29 29 1
695520 46 46 0
726180 39 39 0
753480 46 45 0
950976 27 27 0
1089270 42 42 0
1421280 47 46 0
1539720 47 46 0
PS 这无关紧要,但如果有人能指出为什么我的输出在以 672 开头的行之后发生了变化,我将非常感激。
链接到对应问题的材料。
http://planetmath.org/OreNumber;
http://steve.hollasch.net/cgindex/coding/ieeefloat.html
http://www.cs.yale.edu/homes/aspnes/pinewiki/C(2f)FloatingPoint.html
https://en.wikipedia.org/wiki/Single-precision_floating-point_format
https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html