for m=j/i*(n-1)+1:在matlab里 像2FSK 2PSK 等里都有用到

for(j=1;j<=n;j++) { for(i=1;i<m;i++) 数据结构,怎么循环 main() {long m,n,i=1,j,s; scanf(%ld,%ld,&m,&n); for(;i #include main() { int i,j,m=1; for(i=1;i0;j--) { if(i*j>3) break; m*=#include main(){ int i,j,m=1;for(i=1;i0;j--){ if(i*j>3) break;m*=i*j;}}printf(”m=%d ”,m);}我得出12, for m=j/i*(n-1)+1:在matlab里 像2FSK 2PSK 等里都有用到 Dim j%,i%,m% m = 0 For i = 1 To 3 For j = 1 To i m = m + j Next j,i Text1 = Str(m)最后m j 是多少 #includemain(){int i,j,m=1;for(i=1;i0;j--){if(i*j>3)break;m*=i*j;}}输出m 2个矩阵相加,For i = 1 To 4For j = 1 To 4c(i,j) = a(i,j) + b(i,j)Picture3.Print c(i,j);If i = 1 And j = 1 ThenMax = c(i,j)ElseIf Max < c(i,j) Then Max = c(i,j):n = i:m = jEnd IfNext jPicture3.PrintNext i这个和For i = 1 To 4For j = 1 To 4c(i,j) matlab for循环嵌套语句,帮我看一下错在哪里公式是C(m,n)=-sum(abs(s(i,：)^4*exp(-j*(m-n)*2*pi*sin(rtheta(i))/l)); 程序这么写,错在哪里?for m=1:sensor_number;for n=1:sensor_number;i=1:ll;C(m,n)=-sum(abs(s(i,：)^4*exp(-j*(m-n)* 最大公约数和最小公倍数的程序：#includevoid main(){int m,n,i=1,j,s;scanf(%d,%d,&m,&n);for(i=1;i mmain() {int i,j,m=1; for(i=1;i0;j--) {if(i*j>3) break; m*=i*j; } } printf(%d, 下面这个程序有什么错?为什么运行出来的z(i,j)总是为0呢? lingosets:m/1..6/:;n/1..5/:;ajs(m,n):x,y,z；endsetsmax=@sum(m(i):@sum(n(j)：y(i,j)*z(i,j)+x(i,j);@for(ajs:@bin(z));@for(m(i):z(i,1)+z(i,2)+z(i,3)+z(i,4)+z(i,5) #include main() {inti,j,m=1; for(i=1;i0;j--) {if(i*j>3)break; m*=i*j; } k(i,j)(m)=k(i,j)(m-1)-k(m,i)(m-1)*k(m,j)(m-1)/k(m,m)(m-1) 在matlab中怎么实现啊,谢谢各位大虾!k(i,j)(m)=k(i,j)(m-1)-k(m,i)(m-1)*k(m,j)(m-1)/k(m,m)(m-1) 其中k表示矩阵,k(i,j)表示矩阵的某个元素,后面的（m）或（m-1）表 用matlab作拉格朗日插值,所编的函数为：（问题补充处）m=31*24;n=31*4;l=ones(m,n);y2=ones(1,m);%初始化for i=1:m;for j=1:n;s1=1;for k=1:n;if(j~=k)s1=(x(i)-t(k))/(t(j)-t(k))*s1;endendl(i,j)=s1;%计算插值基函数ends=0;for j=1:n matlab多元函数求最小值问题matlab求r =1/(9-j)*(i^2+81-18*j+j^2)^(1/2)*(38/5-19/20*fix(j)+1/20*(8-fix(j))^2)+(fix(j)+1-j)/(9-j)*(i^2+81-18*j+j^2)^(1/2)*(9/5-1/10*fix(j))+(j^2-2*j*n+n^2+m^2-2*m*i+i^2)^(1/2)/(j-n)*(-19/20*fix((n*i-m*j)/(i+n-j- 下面问题if(i == j || i+j == M-1) 中的i+j == M-1看不懂.谁能解释下#include#define M 3 main(){ int i, j, sum; int angs[M][M]; printf(请输入矩阵: ); sum = 0; for(i = 0; i 输出m到n之间的素数#includevoid main(){int i,j,m,n,mark=1,k=0;scanf(%d %d,&m,&n);for(i=m;i #include int main() { int a[150];int m,n,i,j,t; while(scanf(%d %d,&m ,&n)!=EOF) {printf(#includeint main(){int a[150];int m,n,i,j,t;while(scanf(%d %d,&m ,&n)!=EOF){printf(%d~%d prime include:,m,n);t=0;for ( i=m+1;i