输如任意整数n,调用函数sum(int n),完成对该整数n各位数字求和

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

用递归函数#include#includeintinv(intx);main(){intsum=0,i;for(i=1;i

#includevoidsx(intx[],intn){inti,j,t;for(i=0;i

#include<stdio.h>int getmax(int a,int b){      re

#includevoidswap(int*,int*);voidmain(){inta,b;scanf("%d%d",&a,&b);swap(&a,&b);printf("%d%d\n",a,b);}

intT(intn,intm){\x05if(m>10||m\x05intx=n;\x05intk;\x05intresult=0;\x05intpowNum=1;\x05do{\x05k=x%10;

voidmain(){inti,n;scanf("%d",&n);for(i=0;i*i

VC++编程#includeusingnamespacestd;intmax(intx,inty);//子函数的声明,子函数的参数的类型intmain(){intMAX,x,y;cin>>x>>y;M

如果需要n是变量,则使用指针,以下是原地转置voidTrans(int*a,intn){inttemp,i,j;for(i=0;ifor(j=0;j{temp=a[i*n+j];a[i*n+j]=a[

sum(n){intsum,i;if(i=1;i

#includeintStrchr(chars[],charch){intl,i;intj=0;l=strlen(s);for(i=0;i

第1次:factorial(5)sum+=5*factorial(4)第2次:factorial(4)sum+=4*factorial(3)第3次:factorial(3)sum+=3*factori

#includedoublesum(doublex,intn){\x05doubles=1;\x05inti;\x05doublet=1;\x05doublea=1;\x05for(i=1;i再问：以

有可能除出小数的,都要定义为浮点型,i的步长要增加,正确的代码：#include"stdio.h"floatmath(inti,intn){floats=0;floatk=1;while(i

#include"stdio.h"longintcal(longinta){for(longinti=1,sum=1;i

源码如下：#includeintsum(inta,intb){\x09returna+b;\x09}intmain(){\x09intm=1,n=2,k=3;\x09intt=sum(sum(m,n)

intsum(intn){if(0>n){n=-n;}intsum=0;while(0!=n){sum+=n%10;n/=10;}returnsum;}

clearallclcdt=0.01;t=[0:dt:1.5];w=4*sqrt(3);y=sqrt(3)/2*exp(-4*t).*sin(w*t+pi/3);Dy=diff(y)/dt;subpl

java版实现public class Calculator{    public int calTest(int&nb

#includeintGcd(intM,intN){intRem;while(N>0){Rem=M%N;M=N;N=Rem;}returnM;}voidmain(){inta,b