编写函数fun,求任意m的n次方

intfun(intn){inta=n,b=0;while(a>0){b=b*10;b=b+a%10;a=a/10;}printf("%d",b);getch();return0;}或者把后三行删掉,

#includelongfactorial(intm,intn){longsum=1,sum1=1;inti;if(m-n>n){for(i=m;i>m-n;i--)sum*=i;for

#includeintsum(intn){ints=0;while(n){s+=n%10;n/=10;}returns;}intmain(void){ints=0;for(int

floatfun(intm){    inti,j,count,iszishu;    floatsum; &

#includeintfun(intn){intsum=1;intsuma=0;for(intj=1;j

intfun(intn){intm=1,sum=0,i,j;for(i=1;i

#includeintfac(intx){intret=1,i;for(i=1;i

functionzanswer=cali(n)count=0;fori=0:ncount=count+2^i;endzanswer=count;不好意思,看错了,不知你的代码怎么敲的functionz

voidfun(int*w,intn,intm){\x05inti;\x05intj;\x05inttemp;\x05for(i=0;i0;j--)\x05\x05{\x05\x05\x05w[j]=

longpower(intm,intn){doublep=1;if(n>0){p=m*power(m,(n-1));returnp;}}voidmain(){intm,n;longk;scanf("%

voidfun(longn){intwan,qian,bai,shi,ge;//定义各数位longnixushu;//定义逆序数wan=n/10000;qian=(n-wan*10000)/1000;

intfun(intn){inta=n,b=0;while(a>=1){b=10*b+a%10;a=a/10;}returnb;}已经调试通过了哦!

/*是p=m!/n!(m-n)!*/floatfun(intm,intn){floatp,t=1.0;inti;for(i=1;i

Functions(ByValaAsInteger,ByValbAsInteger)AsIntegerDimiAsIntegerFori=1To1000IfiModa=0AndiModb=0Thens

%编成M函数文件运行后,在命令窗口输入要知道的自然数n,即可求得对应项的Fibonacci数列%有哪步有疑问请问user_entry=input('Pleaseenterthenumberyouwan

用这肯定可以intfun(intm){for(inti=m-1;i>1;i--){intj;booleanisPrime=true;for(j=2;jif(i%j==0){isPrime=false;

/*Note:YourchoiceisCIDE*/#include"stdio.h"voidmain(){intfun(inta[50][50],intm,intn);intm,n,i,j,a[50]

functionfun(d,h){if(d

要求三阶导数,只要写入以下命令即可：clc;clear;diff('sin(x)+x*+x*exp(x)poly函数是根据参数返回一个多项式的功能.

functionf=d(n)f(1)=1;f(2)=1;fori=3:nf(i)=f(i-1)+f(i-2);end