int a=0;after excution of statement a=(2*3,3*4,4*5,5*6);the value of a is?

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inta=0;afterexcutionofstatementa=(2*3,3*4,4*5,5*6);thevalueofais?inta=0;afterexcutionofstatementa=(2

int a=0;after excution of statement a=(2*3,3*4,4*5,5*6);the value of a is?
int a=0;after excution of statement a=(2*3,3*4,4*5,5*6);the value of a is?

int a=0;after excution of statement a=(2*3,3*4,4*5,5*6);the value of a is?
the value of a is 30

C++ 用类解决毕达哥拉斯三元组#includeusing std::cout;using std::endl;class functionBDGLS{public:functionBDGLS( int,int,int );int BDGLS( int,int,int );private:int a;int b;int c;}functionBDGLS::functionBDGLS( int ,int ,int ){int a = 0;int b 合并排序 #includestdio.hvoid merge(int*a,int p,int q,int m){int t[20];int k[20];int n1=q-p+1;int n2=m-q;for(int i=0;i int a=0;after excution of statement a=(2*3,3*4,4*5,5*6);the value of a is? int*a[3] for(int i=0;i>ba[i]=new int[b] } void fun(int *a,int n) { int i,j,k,t; for(i=0;i #include void fun(int a[],int n) { int i,t; for(i=0;i 代码如下:a.h文件extern int p_abs(int &a);extern int p_max(int &a,int &b);class a{public:a():x(0),y(0){}a(const int &i,const int &j):x(i),y(j){}int get_x(){return x;};int get_y(){return y;};void set_x(const int &i){x=i;};void set_y(const int & #include class A { int a; public:A(int aa=0) { a=aa; } A(){cout #include   int inc(int a)   {    return(++a);   }   int multi(int*a,int*b,int*c)   {    return(*c=*a**b);   }   typedef int(FUNC1)(int in);   typedef int(FUNC2) (int*,int*,int*);   void show(FUNC2 fu int a[10][11]={{0, int a = 0啥意思啊 error C2065:'a' :undeclared identifier#includeint main(){int print_shang();int print_xia();print_shang();print_xia();return 0;}int print_shang(){int a,b;for(a=1;a int a=-100, int a=2, 矩阵相加(C++)#include using namespace std; const int rows=3;const int cols=3;void matrixadd(int *,int *,int *,int,int);int main(){int a[rows][cols]={{1,3,5},{7,8,11},{13,15,17}};int b[rows][cols]={{9,8,7},{6,5,4},{3,2,1}};int c[rows][cols]={0 我想搞个X的Y次方的 算法 #include stdafx.hint main(int argc,char* argv[]){int pow(int x,int y);int a,b,c;scanf(%f,%f,&a,&b);c=pow(a,b);printf(%f ,c);return 0;}int pow(int x,int y){int i,z;i=1;z=x;while(i 英语翻译int a;std::cin>>a;int b;std::cin>>b;int value=a;int pow=b;int result =1;for (int cnt=0;cnt =pow;++cnt)result*=value;std::cout 一个关于C语言指针的问题,源程序:#include intmain(){int a[5]={1,2,3,4,5};int *ptr1=(int *)(&a+1);int *ptr2=(int *)((int)a+1);printf(%x,%x,ptr1[-1],*ptr2);return 0;}