一个数的1024 末尾有多少个0三个零兄读一个零

1乘2乘3乘……乘60的积末尾有几个零_百度知道
1乘2乘3乘……乘60的积末尾有几个零
有9个500尾数中10,有93个100,有92个5,有9个50
其他类似问题
其他32条回答
题目就是1到927真百整十的数的零个数的和 92+9=101个 只有相成的因数里有零,积里才有零呀 事实上是83个整十数和9个整百数 所以83+9*2=101
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1乘2乘3乘……乘60的积末尾有14个零
[提问者采纳]
(N /5 )+ (N/5^2)+ (N / 5^3]……其中,N=927. Log5,X≈927.
1.1到927整百整十的数的零个数的和92+9=1012.2*5=10
93101+93=194
huwentingtony的算法是对的!204个!!!!
去掉下面重复的
每个造成0的数目0
42,4,8,16与上面相成可得尾数为0,这类数多的足够了,不计.故共有:99+74+12*2+6*3+1*4=219(也可由第2列直接加99+93+19+7+1=219)
应该是204个零,而不是214个。
10-920共92个个为数零 100-900共9个十位数零 5-925共93个个位数5这些5随便乘以一个偶数就多一个0 50-850有9个十位数5 500有一个百位数5 一共 92+9+93+9+1=204个吧 还是漏了 25,75,125,175,…,925共19个数可以再多一个0 在此基础上 125,375,625还可以再多一个 再在此基础上,625还有多的一个0 250,750分别多一个0所以204+19+3+1+2=229剩一个不知道少在那里楼上的答案比较好
50-850有9个十位数5,少了900这一个。一共230个。
[927/5]+[927/25]+[927/125]+[927/625]=185+37+7+1=230简单的数论题!
参考资料:
在1~927中5的倍数有185个:5,10,...,925. 在1~927中5^2的倍数有37个:25,50,...,925. 在1~927中5^3的倍数有7个:125,250,...,875. 在1~927中5^4的倍数有1个:625. 将1*2*3*4*......*927中的所有的5提出,一共有185+37+7+1=230个5相乘. 1*2*3*4*......*927中的所有的2提出,可见2的数量大于5的数量. 【例如:1~9中只有一个5,而2有7个(&2&中一个、&4&中两个、&6&中一个、&8&中三个)】 而只有2*5个位得零 故1*2*3*4*......*927的得数后面共有230个零.所以选C.
选C我用C++做的:源程序://1乘2乘3一直乘到num,求结果末尾有多少个连续的0&nbsp#include&nbsp&iostream&using&nbspnamespace&class&nbspSTACK{public:&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspvoid&nbsppush(int&nbspe);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspint&nbsppop(void);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspvoid&nbspinit(int&nbspnum);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspvoid&nbspdestroy(void);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspvoid&nbspshow(void);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspint&nbspgetlen(void);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspint&nbspgettop(void);private:&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspint&nbsp*&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspint&nbsp*&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspint&};void&nbspSTACK::init(int&nbspnum){&nbsp&nbsp&nbsp&nbsp&nbspbottom=(int&nbsp*)malloc(num*sizeof(int));&nbsp&nbsp&nbsp&nbsp&nbsp*bottom=0;&nbsp&nbsp&nbsp&nbsp&nbsptop=bottom+1;&nbsp&nbsp&nbsp&nbsp&nbspmaxlen=}void&nbspSTACK::push(int&nbspe){&nbsp&nbsp&nbsp&nbsp&nbspif&nbsp(getlen()==maxlen)&nbspcout&&&the&nbspstack&nbspis&nbspfull!!&&&&nbsp&nbsp&nbsp&nbsp&nbspelse&nbsp&nbsp&nbsp&nbsp&nbsp{&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp*top=e;&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsptop++;&nbsp&nbsp&nbsp&nbsp&nbsp}}void&nbspSTACK::show(void){&nbsp&nbsp&nbsp&nbsp&nbspint&nbsp*p=bottom+1;&nbsp&nbsp&nbsp&nbsp&nbspint&nbsplen=getlen();&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspcout&&&the&nbspstack&nbsplength&nbspis:&&&len&&&.&&&&nbsp&nbsp&nbsp&nbsp&nbspif&nbsp(len==0)&nbspcout&&&the&nbspstack&nbspis&nbspempty!!no&nbspelements&nbspcan&nbspbe&nbspshowed..&&&&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspwhile&nbsp(p!=top)&nbsp&nbsp&nbsp&nbsp&nbsp{&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspcout&&*p&&&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspp++;&nbsp&nbsp&nbsp&nbsp&nbsp}}void&nbspSTACK::destroy(void){&nbsp&nbsp&nbsp&nbsp&nbspfree(bottom);}int&nbspSTACK::getlen(void){&nbsp&nbsp&nbsp&nbspreturn&nbsptop-bottom-1;}int&nbspSTACK::pop(void){&nbsp&nbsp&nbsp&nbspif&nbsp(getlen()==0)&nbspreturn&nbsp-1;&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsptop--;&nbsp&nbsp&nbsp&nbspreturn&nbsp*}int&nbspSTACK::gettop(void){&nbsp&nbsp&nbsp&nbspint&nbsp*p=top-1;&nbsp&nbsp&nbsp&nbspif&nbsp(getlen()==0)&nbspreturn&nbsp-1;&nbsp&nbsp&nbsp&nbspreturn&nbsp*p;}int&nbspadd0(int&nbspn,STACK&&nbspss){&nbsp&nbsp&nbsp&nbsp//ss.show();&nbsp&nbsp&nbsp&nbspif&nbsp(n==0)&nbspreturn&nbsp0;&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspswitch&nbsp(n%10)&nbsp&nbsp&nbsp&nbsp{&nbsp&nbsp&nbsp&nbspcase&nbsp0:&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspreturn&nbspadd0(n/10,ss)+1;&nbsp&nbsp&nbsp&nbspcase&nbsp2:case&nbsp4:case&nbsp6:case&nbsp8:&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspif&nbsp(&nbspss.gettop()%10&nbsp==5&nbsp)&nbspreturn&nbspadd0(&nbspss.pop()*n,&nbspss);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspelse&nbsp{&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspss.push(n);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspreturn&nbsp0;&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp}&nbsp&nbsp&nbsp&nbspcase&nbsp5:&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspif&nbsp(&nbspss.gettop()%2&nbsp==0&nbsp&nbsp)&nbspreturn&nbspadd0(&nbspss.pop()*n,&nbspss);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspelse&nbsp{&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspss.push(n);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspreturn&nbsp0;&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp}&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspdefault:&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspreturn&nbsp0;&nbsp&nbsp&nbsp&nbsp}}&nbsp&nbsp&nbsp&nbspint&nbspmain(){&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspint&nbspi=1;&nbsp&nbsp&nbsp&nbspint&nbspcounter=0;&nbsp&nbsp&nbsp&nbspint&&nbsp&nbsp&nbsp&nbspSTACK&&nbsp&nbsp&nbsp&nbspss.init(1000);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspcout&&&┏━━━━━━━━━━━━━━━━━━━━━━━┓&&&endl&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&&&┃1乘2乘3一直乘到num,求结果末尾有多少个连续的0&nbsp┃&&&endl&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&&&┗━━━━━━━━━━━━━━━━━━━━━━━┛&&&&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspcout&&&请输入num的值:&;&nbsp&nbsp&nbsp&nbspcin&&&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspwhile&nbsp(i&=num)&nbsp&nbsp&nbsp&nbsp{&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspcounter+=add0(i,ss);&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspi++;&nbsp&nbsp&nbsp&nbsp}&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp//ss.show();&nbsp&nbsp&nbsp&nbspcout&&&共有&nbsp&&&counter&&&&nbsp个0。&&&&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbspss.destroy();&nbsp&nbsp&nbsp&nbspsystem(&PAUSE&);&nbsp&nbsp&nbsp&nbspreturn&nbsp0;}&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp&nbsp
92+9+93+9+1=204
204+19+3+1+2+1=230
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出门在外也不愁读数时,末尾有几个0都只读一个零.______(判断对错_百度知道
读数时,末尾有几个0都只读一个零.______(判断对错
读数时,末尾有几个0都只读一个零.______(判断对错)
提问者采纳
一个数末尾不管有几个零,都只读一个零.是错误的,末尾的零不读.故答案为:×.
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