本文介绍了计算值73!的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我们可以算出73! c或c ++中的值,不会丢失有效数字。

can we calculate the 73! value in c or c++ without loss of significant digits.

推荐答案




是的。我自己做的。


-

/ - Joona Palaste(pa*****@cc.helsinki.fi) - ------------芬兰-------- \

\-- -------------------- - 规则! -------- /



Yes. I''ve done it myself.

--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/





是的。你去吧:


/ * homework.c * /


#include< stdio.h>

#include< string.h>


char f [] =" 4470115461512684340891257138125051110076800700282 9050"

" 1581908009237042210406718331701690368000000000000 0001" ;;


int main(无效)

{

f [strlen(f)-1] - ; / *计算73!从73!+ 1 * /

put(f);

返回0;

}


在c ++中它更容易,因为你可以省略void参数到main。


祝你好运,


Sidney



Yes. Here you go:

/* homework.c */

#include <stdio.h>
#include <string.h>

char f[] = "4470115461512684340891257138125051110076800700282 9050"
"1581908009237042210406718331701690368000000000000 0001";

int main(void)
{
f[strlen(f)-1]--; /* calculate 73! from 73!+1*/
puts(f);
return 0;
}

In c++ it''s even easier, since you can omit the void parameter to main.

Best regards,

Sidney




位数。


奇怪的是,口袋科学计算器的范围大于

C的长度或甚至是双精度数。你会注意到计算器不会计算73!它可以完成,但它需要编码(或获取)一种处理非常大数字的方法。

的人问这个措辞相当措辞的人可能希望

答案是不,但我不是一个心灵读者。


简而言之,我们*可以*做几乎任何我们可以在

中表达的数学问题。或者任何我们可以通过

数学运算序列来表示的方式。


digits.

Oddly enough, the range of a pocket scientific calculator is greater than
that of C for long or even double precision numbers. And you will note that
the calculator will not compute 73!. It could be done but it would require
coding (or acquiring) a method of dealing with very large numbers. The
person who asked that rather poorly worded question probably expects the
answer to be "no", but I am not a mind reader.

In short, we *can* do almost any mathematical problem we can formulate in
our mind. Or anything that we can conjure up a way to represent by a
sequence of mathematical operations.


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11-01 01:54