定义一个函数digit(n.k),它送回整数n的从右边开始数第K个数字的值

#includeintdigit(intn,intk){returnk>1digit(n/10,k-1):n%10;}intmain(){printf("%d",digit(12345,3));}

#include"stdio.h"intfact1(intn){if(n==1||n==0)return1;elsereturnn*fact1(n-1);}intfact2(intn){inti,t=

#includeintdigit(intx,intn){intret=0,i;for(i=1;i

根据已知条件有f(1)·f(2)·…·f(k)=log2（3）*log3（4）*log4（5)*logk+1(k+2)=log2（3）/log4（3）*log4（5）*logk+1(k+2)(换第二项

functioncountdigit(number,digit){vartmp=(number+"").replace(/[^2]/ig,"");returntmp.length;}

c++写的#include#include#includeusingnamespacestd;voiddigit(intn,intk){stringstr;char*my;intlen=0;itoa(

类形不匹配错误.改正后的程序：programzlx52;varn,k:longint;functiondigit(n,k:longint):longint;vara,b:longint;begina:

digit(char*n,intk){intn_len=strlen(n);/*对n_len进行大小验证自己加*/chara=*(n+n_len-k);}

具体代码如下：#includeintcountdigit(intnumber,intdigit){intcount=0;while(number){if((number%10)==digit)coun

楼下:现在对了.分段法.an奇(=f(1)+f(3)+...+f(2^n-1)=2^(2n-2)(好化简),an偶=f(2)+...+f(2^n)=1/2(2^(2n-2)+2^(n-1)).所以an

sum=1*k+2*k=#^GDGHNJI(^@~#$^&I(%$EUIO)(max)GJXDT$^($##@~*(min)

typedefintDataType;intN=0;DataType*a;//在需要用到a[N]时,(C++)申请空间a=newa[N];(C语言)申请空间a=malloc(sizeof(DataTy

#includeintsum_of_square(intn){\x09intsum=0;\x09if(n

#include#include"string.h"/*函数功能：返回数字num从左数起第k位.当k非法时（如k

#include#includeintdigit(intn,intk){if(n

在matlab里面：文件->新建->函数functionm=f(n)m=zeros(n,n);fork=1:nm(k,:)=[k-1:-1:10:n-k];endend

#include#include//请自己判断异常情况intdigit(intnum,intk){\x05returnnum/(int)pow(10,int(log10(num))+1-k)%10;}

#include<stdio.h>intdigit(intn,intk){while(--k){n/=10;//右移}returnn%10;//返回个位}voidmain(){intn,k

#include#include#includeusingnamespacestd;voiddigit(intn,intk){stringstr;char*my;intlen=0;itoa(n,my,

#includeintdigit(intn,intk){\x09inti,temp1,temp2;\x09for(i=0;i\x09{\x09\x09temp1=n%10;\x09\x09n=n/10