英语翻译for k = 1:12N = X(n,:) + X(s,:) + X(:,e) + X(:,w) + X(n,e) + X(n,w) + X(s,e) + X(s,w)RAND = rand(m)X = X | (N.*RAND>0.99)endm = 50;X = zeros(m,m);X(25,25) = 1;n = [m 1:m-1];e = [2:m 1];s = [2:m 1];w = [m 1:m-1];
来源:学生作业帮助网 编辑:六六作业网 时间:2024/11/25 14:35:43
英语翻译fork=1:12N=X(n,:)+X(s,:)+X(:,e)+X(:,w)+X(n,e)+X(n,w)+X(s,e)+X(s,w)RAND=rand(m)X=X|(N.*RAND>0.99)
英语翻译for k = 1:12N = X(n,:) + X(s,:) + X(:,e) + X(:,w) + X(n,e) + X(n,w) + X(s,e) + X(s,w)RAND = rand(m)X = X | (N.*RAND>0.99)endm = 50;X = zeros(m,m);X(25,25) = 1;n = [m 1:m-1];e = [2:m 1];s = [2:m 1];w = [m 1:m-1];
英语翻译
for k = 1:12
N = X(n,:) + X(s,:) + X(:,e) + X(:,w) + X(n,e) + X(n,w) + X(s,e) + X(s,w)
RAND = rand(m)
X = X | (N.*RAND>0.99)
end
m = 50;
X = zeros(m,m);
X(25,25) = 1;
n = [m 1:m-1];
e = [2:m 1];
s = [2:m 1];
w = [m 1:m-1];
英语翻译for k = 1:12N = X(n,:) + X(s,:) + X(:,e) + X(:,w) + X(n,e) + X(n,w) + X(s,e) + X(s,w)RAND = rand(m)X = X | (N.*RAND>0.99)endm = 50;X = zeros(m,m);X(25,25) = 1;n = [m 1:m-1];e = [2:m 1];s = [2:m 1];w = [m 1:m-1];
程序语言 可以达到你的点击 放大效果
C语言for(n=k;1
#include int main() { int n,a,sum=1,k; scanf(%d,&n); for(k=n;k>0;k--) sum=sum*k; // a=su
int i,j,k,n; for(n=0;n
#include main() { int i,k; for(i=0,k=-1;k=1;k++) printf(****
); }
(n+1)!/k!- /(k-1)!=(n+1)!/k!- k*n!/k*(k-1)!=(n+1)!/k!- kn!/k!=[(n+1)!-kn!]/k!=(n-k+1)n!/k(n+1)!/k!- /(k-1)!=(n+1)!/k!- k*n!/k*(k-1)!=(n+1)!/k!- kn!/k!=[(n+1)!-kn!]/k!=(n-k+1)n!/k!k*n!/k*(k-1)!怎么等于kn!/k!
请问这三个语句是什么意思?谢谢啦matlab 中 R = zeros(N,N); for (k=1:K) R(k,k) = 1;
for(i=0,k=-1;k=1;k++) printf(*****
);是先判断还是先执行?
请问1^k+2^k+3^k+.+n^k=?
c语言 求1^k+2^k+3^k+……+n^k,假定n=6,k=4#includeint sum(int n,int k){int i;int s=0;for(i=1;i
证明:(n+1)!/k!-n!/(k-1)!=(n-k+1)*n!/k!(k≤n)
VFP题,N=10 S=0 FOR K=N-8 TO N-9 S=S+K N=N-1 ENDFOR S RETURN我不明白K=N-8 TO N-9 ,谁能帮我理解下
英语翻译include stdio.hvoid main (){int i,j,k,n;printf('water flower 'number is :);for(n=100;n
for(;(k%m!=0) || (k%n!=0);k++)什么意思
#include #include fun(int n) { int k,r; for(k=2;k
void fun(char *a1,char *a2,int n) { int k; for(k=0; k
Matlab LU分解程序,总是说括号不对称或异常function [l,u]=lu_fenjie(A)n=length(A);u=zeros(n);l=eye(n);u(1,:)=A(1,:);l(2:n,1)=A(2:n,1)/u(1,1);for k=2:nu(k,k:n)=A(k,k:n)-l(k,1:k-1)*u(1:k-1,k:n);l(k+1:n,k)=(A(k+1:n,1:k-1)*u(1:k-1,k))/u(k,k);
试证明 x/[n(n+k)]=(x/k)[1/n-1/(n+k)]
关于一道数据结构计算时间复杂度的问题例题如下:FOR i:=1 TO n DO ----------{n+1} FOR j:=1 TO n DO ----------{n*(n+1)} [ c[i,j]:=0; -------------{n的2次方} FOR k:=1 TO n DO ----------{n的2次方*(n+1)} c[i,j]:=c[i,j]+a[i,k]*b[k,