#include #define N 2#define M N+1#define NUM (M+1)*M/2main()
在C语言中为什么执行过 #define N 2 #define M N+1 #define NUM (M+1)*M/2以
#include #define X 5 #define Y X+1 #define Z Y*X/2 main() {
#define ADC_CHSEL_RES(n,m) _SBF(4*n,m) #define ADC_INT_ENABL
#include #define M 20; void main() { int i,j,k,p,t,n=0; int
#define N 5 #define f(M) ((N+1)*M) 求x=2*(N+1)+2*f(N+1); 求x的值
define M 5#define N M+M main(){int k;k=N*N*5; printf("%d",k)
#include #include #define TRUE 1 #define FALSE 0 #define OK
C语言计算小问题#include#define Add(x) x+xint main(){\x05\x05int m=1
#include #define S(x) x*x void main() { int a,k=3,m=1; a=S(k
#define SQR(X) X*X #include void main(){ int a=16,k=2,b=4,m=
#define N 3 #define Y(n) ( (N+1)*n) 则执行语句:z=2 * (N+Y(5+1));后
#define N 3 #define Y(n) ((N=1)*n) 则表达式2*(N+Y(5+1))的值是