int x=3;main(){int i;for(i=1;i
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intx=3;main(){inti;for(i=1;iintx=3;main(){inti;for(i=1;iintx=3;main(){inti;for(i=1;i百度
int x=3;main(){int i;for(i=1;i
int x=3;main(){int i;for(i=1;i
int x=3;main(){int i;for(i=1;i
百度
C语言问题int x=3;incre();main(){ int i; for (i=1;i
int x=3;main(){int i;for(i=1;i
int x=3;main(){int i;for(i=1;i
main(){int x,i;for(i=1;i
main(){ int i=1;while (i
#include using namespace std; int main() { int a,b,c; a=3; int f(int x,int y,int z);#include using namespace std; int main() {int a,b,c;a=3;int f(int x,int y,int z); cin>>a>>b>>c;c=f(a,b,c);cout
main(){ int x=5; while(x
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
麻烦不要复制百度上的答案,int fa(int x){ return x*x; }int fb(int x){ return x*x*x; }int f(int (*f1)(),int (*f2)(),int x){ return f2(x)-f1(x); }main(){ int i;i=f(fa,fb,2); printf(“%d
”,i);}
c语言求出 2 到 m 之间 素数#include int main (){int i,m,x=1,j;scanf(%d,&m);for(i=3;i
求此题解题过程:#include #define N 8 void fun(int *x,int i) {*x=*(x+i);} main() {int a[N]
求高手解释一下这段程序#include class Test {int x,y; public: Test(int i,int j=0) {x=i;y=j;} int get(int i,int j) {return i+j;} }; void main() {Test t1(2),t2(4,6); int (Test::*p)(int,int=10); p=Test::get; cout
(36)有以下程序main(){int x[3][2]={0},I;for(I=0;I
正确的应该怎么编,还有具体的解释,#includeclass Coord {public:void setCoord(int a,int b){ x=a; y=b; }int getx(){ return x; }int gety(){ return y; }private:int x,y;};void main(){ Coord op1;int i,j;op1.x=3;op1.y=4;op1.setCoord(5,6); // 调
#include int b=2; int fun(int*k) { b=*k+b;return(b);} main() {int a[10]={1,2,3,4,5,6,7,8}#include int b=2; int fun(int*k) { b=*k+b;return(b);} main() {int a[10]={1,2,3,4,5,6,7,8},i;for(i=2;i
void main() {int x,i,n=0; for(x=100;i
void main() {int x,i,n=0; for(x=100;i
#define N 20 fun(int a[],int n,int m) {int i; for(i=m;i>n;i--)a[i+1]=a[i]; return m; } void main() #define N 20fun(int a[],int n,int m){int i;for(i=m;i>n;i--)a[i+1]=a[i];return m;}void main(){ int i,a[N]={1,2,3,4,5,6,7,8,9,10};fun(a,0,N/2);for(i=0;i