I=｛1,2,3,4,5,6,7,8,9｝A≦I,B≦I,A∩B=｛2｝,（CuA）∩（CuB）=｛1,9｝,CuA∩B=｛4,6,8｝求A∩（CuB）

#define N 20 fun(int a[],int n,int m) {int i; for(i=m;i>n;i--)a[i+1]=a[i]; return m; } void main() #define N 20fun(int a[],int n,int m){int i;for(i=m;i>n;i--)a[i+1]=a[i];return m;}void main(){ int i,a[N]={1,2,3,4,5,6,7,8,9,10};fun(a,0,N/2);for(i=0;i 试求i^1,i^2,i^3,i^4,i^5,i^6,i^7,i^8i是虚数由题目推测出i^n的值的规律,用式子表达 I 1 3 0 I I 7 I I 0 1 -1 I X= I 2 I求X.大哥大姐,用矩阵的初等变换写~I 2 1 5 I I 4 I 计算；（1）,（-8-7i)(-3i) (2),(4-3i)(-5-4i) (3),2i/2-i (4),2+(4).2+i/7+4i 下列程序的输出结果是#define N 20void fun (int a[ ],int n,int m){int i,j;for (i=m;i>n;i--)a[i+1]=a[i];}main(){int i,a[N]={1,2,3,4,5,6,7,8,9,10};fun(a,2,9);for(i=0;i for循环中,3重循环,0和1是怎么回事?输出结果：$i=0$i=1零 $i=2零一 $i=3零一二 $i=4零一二三 $i=5零一二三四 $i=6零一二三四五 $i=7零一二三四五六 $i=8零一二三四五六七 $i=9零一二三四五六七八 $i=10 #define N 20 fun(int a[],int n,int m) { int i; for(i=m;i>=n;i--) a[i+1]=a[i]; return a[a+1]; } main#define N 20fun(int a[],int n,int m){ int i;for(i=m;i>=n;i--)a[i+1]=a[i];return a[a+1];}main(){int i,a[N]={1,2,3,4,5,6,7,8,9,10};fun(a,2,9);for(i=0;i 几道高二数学题(5-3i)+(7-5i)-4i(5-3i)+(7-5i)-4i(-2-4i)-(-2+i)+(1+7i)(-2-3i)(-5+i)(1+i)(2+i)(3+i)(3-i)z=4+2i 最好 有比较清楚的过程 [m n]=size(I)subplot(2,1,1),imshow(I);title({'原图像'});y1=I(4,:);y2=I(5,:);y3=I(6,:);y4=I(7,:);y5=I(8,:);y6=I(9,:);y7=I(10,:);y8=I(20,:)*2;subplot(212)plot(1:n,y8) 改写下面的句子,用HE作句子的主语1 I must call the doctor.2 I am going to telephone him.3 I can go with her .4 I have a new car.5 I come from America.6 I am America.7 I like ice cream.8 I want a newspaper.9 I was at school yesterday.10 I main(){int a[3][3]={1,2,3,4,5,6,7,8,9},i=0,sum=0;do{sum+=a[i][i];i++;}while(i (1-3i)-(2+5i)+(-4+9i)等于? (1+7i)^2=?(3+2i)*(3-2i)*(-4+i)=?根号3i分之1=?8+3i分之5-2i=?1-2i分之1+2i - 1+2i分之1-2i=? Dim a, i% a=array(1,2,3,4,5,6,7) For i =Lbound(a) to Ubound(a) a(i)=a(i) *a(i) next i print a(i)下标越界了,为什么啊 main( ) { int i,x[3][3]={9,8,7,6,5,4,3,2,1},*p=&x[1][1]; for(i=0;i main() { inta[3][3]={1,2,3,4,5,6,7,8,9},i,s=1; for(i=0;i #include void fun(int*a) {a[0]=a{1};} main() {int a[10]={10,9,8,7,6,5,4,3,2,1},i;for(i=2;i>=0;i--)fun(&a[i]);for(i=0;i main( ) { int i,x[3][3]={1,2,3,4,5,6,7,8,9}; for(i=0;i