#define N 2 #define M N+1 #define NUM (M+1)*M/2 main() {int I; for(I=1;I

#define #define N 2 #define M N+1 #define K M+1*M/2#define N 2#define M N+1#define K M+1*M/2main(){int i;for(i=1;i #define N 3 #define Y(n) ((N=1)*n) 则表达式2*（N+Y（5+1))的值是#define N 3 #define Y(n) ((N+1)*n) 则表达式2*（N+Y（5+1))的值是 #define N 2 #define M N+1 #define NUM (M+1)*M/2 main() {int I; for(I=1;I #include #define N 2#define M N+1#define NUM (M+1)*M/2main(){ int i;for(i=1;i 在C语言中为什么执行过 #define N 2 #define M N+1 #define NUM (M+1)*M/2以后,NUM的值是8而不是6 #define TURE 1 #define FALSE 0 #define OK 1 #define ERROR 0 #define INFEASIBLE -1 #define OVERFLOW #define TURE 1#define FALSE 0#define OK 1#define ERROR 0#define INFEASIBLE -1#define OVERFLOW -2typedef int status;typedef int ElemType;#include#define #define PWM_PCR_PWMENAn(n) ((uint32_t)(((n&0x7) #define OK 1 #define ERROR 0 #define OVERFLOW -2这些定义有什么作用 #define N 3 #define Y(n) ( (N+1)*n) 则执行语句:z=2 * (N+Y(5+1));后,z的值为 #define getbit( b,n) (((b) & (1L 下面的程序结果是什么?怎么算?急求答案# include # define M 3# define N M+1# define NN N*N/2void main(){ printf(%d ,NN); printf(%d ,5*NN);} define是什么意思 #define N 5 #define f(M) ((N+1)*M) 求x=2*(N+1)+2*f(N+1); 求x的值 最好带运算过程 高手帮忙看一道二级C的真题~题目是这样的#include #define N 2#define M N+1#define NUM (M+1)*M/2main(){int i,n=0;for(i=1;i #include #define X 5 #define Y X+1 #define Z Y*X/2 main() { int a; a=Y; printf(%d %d ,Z结果为什么是7 5呢,不解, #include #define A 2 #define B(x) x*(A+2) void main() { int a=5; printf(%d ,B(a+1)); } C语言中 #define N 30 #define IFADOB(A,B) ((A)&&(B,0))