int a=1,b=2,c=2,t; while(a

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int a=1,b=2,c=2,t; while(a
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int a=1,b=2,c=2,t; while(a
int a=1,b=2,c=2,t; while(a

int a=1,b=2,c=2,t; while(a
a

void fun(int a,int b) { int t; t=a;a=b;b=t; } main()void fun(int a,int b){ int t; t=a;a=b;b=t; } main() { int c[10]={1,2,3,4,5,6,7,8,9,0}.i; for(i=0;i #include void fun(int a, int b) { int t; t=a; a=b; b=t;main() { int c[10]={1,2,3,4,5,6,7,8,9,0), i; for (i=0; i #include void fun(int a,int b) {int t; t=a;a=b;b=t; } main() {int c[6]={1,2,3,4,5,6},i;for(i=0;i #include stdio.h void fun(int a,int b) { int t; t=a,a=b,b=t; } main() { int c[10]={1,2,3,4,5,6,7#include stdio.hvoid fun(int a,int b){int t;t=a,a=b,b=t;}main(){int c[10]={1,2,3,4,5,6,7,8,9,0},i;for(i=0;i 求助MATLAB符号计算:%声明符号变量syms c w t b a D n T v Q x y;%给已知的变量赋值D=0.038;w=2*pi/T;v=n*w*D/2;%正切值b= (sin (w*t))/(c-cos (w*t));%余弦值a=sqrt (1/ (1+b^2));%正弦值d=sqrt ((b^2)/ (1+b^2));x=int (v*sqrt (1/ (1 int w[3][4]={{0,1},{2,4},{5,8}}; int (*p)[4]=w; 则数值为4的表达式是int w[3][4]={{0,1},{2,4},{5,8}};int (*p)[4]=w;则数值为4的表达式是 .(A)*w[1] (B)*p[1] (C) w[2][2] (D)p[1][1] int w[3][4]={{0,1},{2,4},{5,8}}; int(*p)[4]=w;则数值为4的表达式是_.a.*w[1] b.*p[1] c.w[2][2]d.p[1][1] #includemain(){int a=1,b=2,c=2,t;while(a main() { int a=1,b=2,c=2,t; while(a int a=1,b=2,c=2,t; while(a int a=1,b=2,c=3,t;while (a int t,a=5,b=6,w=1,x=2,y=3,z=4;经过t=(a=w>x)&&(b=y+z)后,变量t,a,b分别为多少? c语言这段程序看不懂int fun(int a,int b,int c){ int t; t=(a>b)?(b>c?b:(a>c?c:a)):((a>c)?a:((b>c)?c:b)); return t; } 计算机C语言7.有以下定义语句double a,b;int w; long c;若各变量已正确赋值,则下列选项中正确的表达式是______.(2分) A.a=a+b=b++ B.w%((int)a+b) C.(c+w)%(int)a D.w=a==b; 请问一下“void main() {int t,a,b,c;a=1;b=2;c=2; while(a 求a,b,c值要过程 int a=1,b=2,c=2,t; while(a 以下程序运行时的输出结果第一行至第四行分别为void swap1(int c[ ]){ int t;t=c[0];c[0]=c[1];c[1]=t;}void swap2(int c0,int c1){ int t;t=c0;c0=c1;c1=t;}main( ){ int a[2]={3,5},b[2]={3,5};swap1(a); swap2(b[0],b[1]);printf(“%d %d % matlab 用变量替换表达式syms w t nsyms T positive;a0=(1/(T/2))*(int(0,-T/2,0)+int(1,0,T/2));a_n=(1/(T/2))*(int(0*cos(n*w*t),-T/2,0)+int(1*cos(n*w*t),0,T/2))b_n=(1/(T/2))*(int(0*sin(n*w*t),-T/2,0)+int(1*sin(n*w*t),0,T/2))fx=a0/2+symsum(a_n*cos(