设一个函数,调用它时,每次实现不同的功能

#includefloatadd(floatx,floaty);floatsub(floatx,floaty);floatmul(floatx,floaty);floatprocess(floatx,

javascript程序：functionjl(a,b){a=a||[0,0];b=b||[0,0];x=Math.abs(a[0]-b[0]);y=Math.abs(a.[1]-b.[1]);ale

intfun(intn){returnn%2;}intmain(){intn;printf("\n输入要判断的整数");scanf("%d",&n);if(fun(n)==0)printf("\n%d

int去掉,调用时候不用类型声明

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[

#includeintfun(inta,intb){return(a+b)*(a+b);}intmain(){inta,b;scanf("%d\n%d",&a,&b);printf("%d\n",fu

其实只要做一次种,然后再调用rand()就能产生不同的结果了,不用每次调用rand()之前都用srand()做种的也就是说你那个程序可以改为srand(time(0));for(a=0;a再问：按您说

#includemain(){intm,b,s,g;for(m=100;m

100x+10y+z=x³+y³+z³

#includefloatmax(floata,floatb){if(a>b)returna;returnb;}voidmain(){floata,b,c,d;scanf("%f,%f,%f",&a,

#includeintf(inta,intb){intcount=0;inti,j;for(i=a;i

#include#defineN9voidmain(){inti,j,a[N][N]={0};for(i=1;i

intm（intx）{if(x==1)return1;elsereturnx*m(x-1);}voidmain(){m(20);}

#includemain(){longintn,i,j,k;doublem;printf("请输入数字'n':");scanf("%d",&n);if(n%2==1){for（i=0;i0){for（

#includevoidswap(int*x,int*y);intmain(void){inta,b;printf("a:");scanf("%d",&a);printf("b:");scanf("%

voids(folata,folatb,folatc){folatp;p=(a+b+c)/2;S=√[p(p-a)(p-b)(p-c)];returns;}

同样的定义,matlab没有像c语言那样严格的参数类型限制你可以定义functionyy=fitnessfct(aa)在函数里面aa(:,n)就是aa的第n列数据aa(m,:)就是aa的第m行数据aa

这个题目应当有原来的程序,让人填空.如果自己写程序,很简单：#include#includevoidfun(int*x,int*y){intz;z=*x;*x=*y;*y=z;}main(){inta

#includeusingnamespacestd;intmain(){booleven(intn);inta,b;cout

第一题：#includevoidmain(){inteven(intn);intm,n;printf("iuputanumber:");scanf("%d",&n);m=even(n);printf(