如下C程序的输出是什么？#include [stdio.h]void Func1 (int x, int y);void Func2 (int *x, int *y); int main() { int x = 3; int y = 4;Func1 (x, y); printf ("x = %d, y = %d\n", x, y);Func2(&x, &y); printf ("x = %d, y = %d\n", x, y);} void Func1 (int x, int y) { x = x + y; y = x - y; x = x - y; printf ("x = %d, y = %d\n", x, y);} void Func2 (int *x, int *y) { *x = *x + *y; *y = *x - *y; *x = *x - *y;;} A: x = 3, y = 4x = 3, y = 4x = 3, y = 4 B: x = 4, y = 3x = 4, y = 3x = 4, y = 3 C: x = 3, y = 4x = 3, y = 4x = 4, y = 3 D: x = 4, y = 3x = 3, y = 4x = 4, y = 3

以下程序的输出结果是（ ）。main(){ int x=1,y=2; void swap(int x,int y); swap(x,y); printf("x=%d,y=%d\n",x,y);}void swap(int x,int y){ x=3,y=4;} A: x=3,y=4 B: x=1,y=2 C: x=3 y=4 D: x=1 y=2

设\(z = {e^ { { y \over x}}} + {x^y} + {y^x}\)，则\({z_x} = \) A: \({1 \over x}{e^ { { y \over x}}} + {x^y}\ln x + x{y^{x - 1}}\) B: \(- {y \over { { x^2}}}{e^ { { y \over x}}} + {x^y}\ln x + x{y^{x - 1}}\) C: \({e^ { { y \over x}}} + y{x^{y - 1}} + {y^x}\ln y\) D: \( - {y \over { { x^2}}}{e^ { { y \over x}}} + y{x^{y - 1}} + {y^x}\ln y\)

下列语句语法正确的是（ ） A: if x<2*y and x>y then y=x**2 B: if x<2*y : x>y then y=x^2 C: if x<2*y and x>y then y=x2 D: if x<2*y and x>y then y=x^2

int x,y; x=2==3-1; y=-5