下面程序的输出结果【9】 。
main ()
{enum team {y1=4,y2,y3};
printf ("%d",y3);}
第1题:
3、第4题博弈树对应的策略式表述,如下表所示 参与人2 (a,c) (a,d) (b,c) (b,d) 参与人1 M -1,-2 -1,-2 x1, y1 x2, y2 N x3,y3 x4,y4 0, 2 1,1 该博弈双矩阵表述的支付值,正确的是()
A.(x1,y1) = (2, 0), (x2,y2) = (2,0), (x3,y3)= (0,2), (x4,y4) = (1,1)
B.(x1,y1) = (-1, -2), (x2,y2) = (-1,-2), (x3,y3)= (0,2), (x4,y4) = (1,1)
C.(x1,y1) = (0, 2), (x2,y2) = (0,2), (x3,y3)= (2,0), (x4,y4) = (2,0)
D.(x1,y1) = (0, 2), (x2,y2) = (0,2), (x3,y3)= (2,0), (x4,y4) = (2,0)
第2题:
分别用红、绿、蓝三种颜色在同一个图形窗口绘制下列函数在区间[-pi, pi]的图形,y=sin(x),y=cos(x),y=tan(x).
A.clear x=-pi:0.1:pi; y1=sin(x); y2=cos(x); y3=tan(x); plot(x,y1,'r',x,y2,'g',x,y3,'b')#B.clear x=-pi:0.1:pi; y1=sin(x); plot(x,y1,'r') y2=cos(x); plot(x,y2,'g') y3=tan(x); plot(x,y3,'b')#C.clear x=-pi:0.1:pi; y1=sin(x); plot(x,y1,'r') hold on y2=cos(x); plot(x,y2,'g') y3=tan(x); plot(x,y3,'b')#D.clear x=-pi:0.1:pi; y1=sin(x); plot(x,y1,'r') y2=cos(x); plot(x,y2,'g') y3=tan(x); plot(x,y3,'b')#E.clear x=-pi:0.1:pi;第3题:
第4题博弈树对应的策略式表述,如下表所示 参与人2 (a,c) (a,d) (b,c) (b,d) 参与人1 M -1,-2 -1,-2 x1, y1 x2, y2 N x3,y3 x4,y4 0, 2 1,1 该博弈双矩阵表述的支付值,正确的是()
A.(x1,y1) = (2, 0), (x2,y2) = (2,0), (x3,y3)= (0,2), (x4,y4) = (1,1)
B.(x1,y1) = (-1, -2), (x2,y2) = (-1,-2), (x3,y3)= (0,2), (x4,y4) = (1,1)
C.(x1,y1) = (0, 2), (x2,y2) = (0,2), (x3,y3)= (2,0), (x4,y4) = (2,0)
D.(x1,y1) = (0, 2), (x2,y2) = (0,2), (x3,y3)= (2,0), (x4,y4) = (2,0)
第4题:
【单选题】下面程序的输出是。 main() {enum team {my,your=4,his,her=his+10}; printf("%d%d%d%dn",my,your,his,her);}
A.0 1 2 3
B.0 4 0 10
C.0 4 5 15
D.l 4 5 15
第5题:
已知输入信号A、B的波形和输出Y1、Y2、Y3、Y4的波形如图所示,试判断各为哪种逻辑门,并画出相应逻辑门图符号,写出相应逻辑表达式。