随机生成5*5的两位整数矩阵,该矩阵的副对角线-搜答案-拍题作业网

fori=1:10000000A=8;B=35;mat=A+(B-A)*rand(10,4)forj=1:10sum(mat(j,:))ifsum(mat(j,:))150;continue;else

C#源程序：class Program{ static void Main()

unifrnd(5,10,20)+diag(inf+zeros(1,20))

是根据对角线对称吗?a=rand(5,5);b=tril(a,-1)+triu(a',0);

Dima%(4,5)PrivateSubCommand1_Click()'产生并输出Fori=1To4Forj=1To5a(i,j)=Int(Rnd*100)'[1,100]Printa(i,j);N

给你个思路吧,对角线是满足I+J=列数-1的所以左右分别是的关系所以三个判断语句就搞定了另外两位偶数是10,调用随即函数要加上这个范伟就行啦

PrivateSubCommand1_Click()DimiAsInteger,myTempNumAsInteger,myTempStrAsStringRandomize(Timer)Fori=1To

#include <stdio.h>#include <stdlib.h>void main(){ int num[5][5]

#include#include#includeusingnamespacestd;constintN=4;intmain(){srand(time(NULL));inti,j,(*a)[N];a=n

PrivateSubCommand1_Click()RandomizeForm1.ClsDimc,eAsIntegerc=0e=0DimdAsDoubled=1Dima(4,4)AsStringFor

10*rand(4,5)

复制如下代码到VBA中,运行即可PublicSubgen()DimiAsIntegerDimjAsIntegerDimkAsIntegerDima(25)AsIntegera(0)=Int(Rnd()

and()%41-20这样吧

random=floor(unifrnd(100,200,5,6))random=19517616114010512012314517919313511916010119219118116014818

#includeusingnamespacestd;constintN=10,M=10;//随机矩阵大小#include#includeintmain(){srand((unsigned)time(N

B=randi([0,99],size(A));常用的分布：Rand是产生0到1的均匀分布；如果要得到在区间[a,b]上的均匀分布,可采用a+（b-a）*rand；Randn是产生标准正态分布；如果想

为了生成某个范围内的随机整数,可使用以下公式：Int((upperbound-lowerbound+1)*Rnd+lowerbound)upperbound是随机数范围的上限,而lowerbound则

m=5;n=5;%矩阵大小m×nA=round(9.*rand(m,n));%5阶矩阵,9的随机整数形成sh=0;xa=0;dia=0;ci=0;kao=0;fori=1:mforj=1:nifj>i

andperm.说我回答太简洁.就是这个函数么.讨厌.

dimstr(4,5)asintegerdimnum1(4),num2(5)asinteger'num1行的和,num2列的和fori=0to3num1(i)=0num2(i)=0nextnum2(4