main(){int n=6,x;x=f(n);printf("%d!=%d\n",n,x);}f(int n){int y'if(n==1)y=1;else y=n*f(n-1)main(){int n=6,x;x=f(n);printf("%d!=%d\n",n,x);}f(int n){int y'if(n==1) y=1;else y=n*f(n-1);return()}求详解 ——————————————— 特

#define f(x) x*x main( ) { int i; i=f(4+4)/f(2+2); printf(%d ,i); } #define f(x) x*x main() {int a=6,b=2,c; C=f(a)?f(b); printf(%d ,c);} 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);} main(){int n=6,x;x=f(n);printf(%d!=%d ,n,x);}f(int n){int y'if(n==1)y=1;else y=n*f(n-1)main(){int n=6,x;x=f(n);printf(%d!=%d ,n,x);}f(int n){int y'if(n==1) y=1;else y=n*f(n-1);return()}求详解 ——————————————— 特 #include void f(int*x,int y) {*x=y+1;y=*x+2;} main() {int a=2,b=2; f(&a,b); printf(%d %d#includevoid f(int*x,int y){*x=y+1;y=*x+2;}main(){int a=2,b=2;f(&a,b);printf(%d %d,a,b);} 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 C语言 答案为什么是8 4 #includevoid f(int y,int *x){y=y+*x;*x=*x+y;}main(){int x=2,y=4;f(y,&x);printf(%d %d ,x,y);} #includevoid f(int y,int *x){ y=y+*x; *x=*x+y;} void main(){ int x=2,y=4;f(y,&x);printf(%d,%d ,x,y);} main(){ int x=5; while(x void f( int y,int *x) {y=y+*x; *x=*x+y;} main( ) { int x=2,y=4; f(y,&x); printf(%d %d ,x,y); } void f( int y,int *x){y=y+*x; *x=*x+y;}main( ){ int x=2,y=4;f(y,&x);printf(%d %d ,x,y);} 执行后输出的结果是 . #includeint a(int x);int b(int y);int main(){int m,n;n=6;m=a(n);printf(%d ,m);return 0;}int a(int x){x=b(x+3);return x;}int b(int y){return y+1;} 有关浮点数的运算程序A：double f(int x){ return 1.0 / x ; }void main(){ double a ,b;int i ;a = f(10) ;b = f(10) ;i = a == b ;printf( %d ,i ) ;}程序B：double f(int x){ return 1.0 / x ; }void main(){ double a ,b ,c;int i ;a = f(10) ;b #include #include #include int fun(float x[],int n) /f#include #include #include int fun(float x[],int n)//if(x[j]=>xa)c++;return ;}main(){float x[100]={193.199f,195.673f,195.757f,196.051f,196.092f,196.596f,196.579f,196.763f};system(cls);printf(%d main( ) {int x=3; do {printf(%d ,x-=2;} while (--x)); } #include main() {double x=1.0; int n; for(n=0;n 下面程序的运行结果是_______#define N 10#define s(x) x*x#defien f(x) (x*x)main(){int i1,i2;i1=1000/s(N);i2=1000/f(N);printf(%d %d ,i1,i2);} #define N 10#define s(x) x*x#define f(x) (x*x)main(){int i1,i2;i1=1000/s(N);i2=1000/f(N);printf(%d %d ,i1,i2);}运行结果是? int fun(static int n) {if(n==1)return 1; else return(n+fun(n-1)); } main() {int x; scanf(%d,&x); int fun(static int n){if(n==1)return 1;elsereturn(n+fun(n-1));}main(){int x;scanf(%d,&x); x=fun(x)+fun(x);printf(%d ,x);}执行程序时,给变量