#define M(x,y,z) x*y+z main() { int a=1,b=2,c=3; printf(“%d\n”,M(a+b,b+c,c+a)); }#define M(x,y,z) x*y+zmain(){ int a=1,b=2,c=3;printf(“%d\n”,M(a+b,b+c,c+a));} 请问怎么计算的?

#define min(x,y) (x #define MIN(x,y)(x) #define X 3 #define Y X*2 #undef X #define X 2 int z=Y; z 的值为多少? #define M(x,y,z) x*y+z main() { int a=1,b=2,c=3; printf(“%d ”,M(a+b,b+c,c+a)); }#define M(x,y,z) x*y+zmain(){ int a=1,b=2,c=3;printf(“%d ”,M(a+b,b+c,c+a));} 请问怎么计算的? #define SETBIT(x,y) (x|=(1 #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, C语言中#define M(X,Y) (X)*(Y)属于什么语法,哪部分知识? 计算数学表达式#include #define FUN(m) (m*m+4.5*m-2.3)/(m-3.5)void main(){ /**/ float x,y,z; /**/ printf( Please input x(3.5),y(3.5),z(3.5): ); scanf(%f%f%f,&x,&y,&z); if(x==3.5 || y==3.5 || z==3.5) { printf(divide 0,error!); #define PR(x,y,z) printf(%d %d %d ,x,y,z) main() { int x,y,z; x=(9+6)%5>=9%5+6%5; printf#define PR(x,y,z) printf(%d %d %d ,x,y,z)main(){int x,y,z;x=(9+6)%5>=9%5+6%5;printf(%d ,x);z=x?(y=x):(y=++x);PR(x,y,z);} #define p(x) x*x int x=5,y=3,z; z=p(x+y); A) 64 B) 23 C) 46 D) 32 # define M(x,y,z) x*y+z main() { int a=1,b=2,c=3; printf(%d ,M(a+b,b+c,c+a)); } # define ABS_MOD(x,y) (((x) < 0) ((((x) % (y)) + (y)) % (y)):((x) % (y))) 阅读下面程序写出执行或输出结果：#define M(x,y,z) x*y+z main() { int a=1,b=2,c=3; printf(%d ,M(a阅读下面程序写出执行或输出结果：#define M(x,y,z) x*y+zmain(){ int a=1,b=2,c=3;printf(%d ,M(a+b,b+c,c+a));} 、 #define M(x,y,z) x*y+z main() { int a=1,b=2,c=3; printf(%d,M(a+b,b+c,c+a)); } 这道题我口算结果、#define M(x,y,z) x*y+zmain(){int a=1,b=2,c=3;printf(%d,M(a+b,b+c,c+a));}这道题我口算结果是19,答案是12,为什么?main(){char a 下列程序段的输出结果是( ).#define M(x,y) x*y printf(%d,M(10+2,2)); (x-y)^2-(z-m)^2因式分解 (x-y)²-(z-m)²因式分解 x-y/(z-x)(z-y) 一y-z/(x-y)(x-z) ＋z-x/(y-z)(y-x)