假设我们有0.33,我们需要输出1/3。 如果有0.4,我们需要输出2/5。
0.33
1/3
0.4
2/5
这样做的想法是使用户易于理解,以使用户理解 y中的 “ x个部分 ”,从而更好地理解数据。
我知道百分比是一个很好的替代品,但我想知道是否有简单的方法可以做到这一点?
我发现David Eppstein 对给定的实数 C代码的有理近似值正是您所要的。它基于连续分数理论,并且非常快速且相当紧凑。
我使用了针对特定分子和分母限制定制的版本。
/* ** find rational approximation to given real number ** David Eppstein / UC Irvine / 8 Aug 1993 ** ** With corrections from Arno Formella, May 2008 ** ** usage: a.out r d ** r is real number to approx ** d is the maximum denominator allowed ** ** based on the theory of continued fractions ** if x = a1 + 1/(a2 + 1/(a3 + 1/(a4 + ...))) ** then best approximation is found by truncating this series ** (with some adjustments in the last term). ** ** Note the fraction can be recovered as the first column of the matrix ** ( a1 1 ) ( a2 1 ) ( a3 1 ) ... ** ( 1 0 ) ( 1 0 ) ( 1 0 ) ** Instead of keeping the sequence of continued fraction terms, ** we just keep the last partial product of these matrices. */ #include <stdio.h> main(ac, av) int ac; char ** av; { double atof(); int atoi(); void exit(); long m[2][2]; double x, startx; long maxden; long ai; /* read command line arguments */ if (ac != 3) { fprintf(stderr, "usage: %s r d\n",av[0]); // AF: argument missing exit(1); } startx = x = atof(av[1]); maxden = atoi(av[2]); /* initialize matrix */ m[0][0] = m[1][1] = 1; m[0][1] = m[1][0] = 0; /* loop finding terms until denom gets too big */ while (m[1][0] * ( ai = (long)x ) + m[1][1] <= maxden) { long t; t = m[0][0] * ai + m[0][1]; m[0][1] = m[0][0]; m[0][0] = t; t = m[1][0] * ai + m[1][1]; m[1][1] = m[1][0]; m[1][0] = t; if(x==(double)ai) break; // AF: division by zero x = 1/(x - (double) ai); if(x>(double)0x7FFFFFFF) break; // AF: representation failure } /* now remaining x is between 0 and 1/ai */ /* approx as either 0 or 1/m where m is max that will fit in maxden */ /* first try zero */ printf("%ld/%ld, error = %e\n", m[0][0], m[1][0], startx - ((double) m[0][0] / (double) m[1][0])); /* now try other possibility */ ai = (maxden - m[1][1]) / m[1][0]; m[0][0] = m[0][0] * ai + m[0][1]; m[1][0] = m[1][0] * ai + m[1][1]; printf("%ld/%ld, error = %e\n", m[0][0], m[1][0], startx - ((double) m[0][0] / (double) m[1][0])); }