编写一个程序实现3*3矩阵的转置,在主函数中用scanf()函数输入以下

我最近在看C++C语言不一定用的很专业见谅啊!#includevoidmain(){intarray[3][4]={0};inti,j,sun=0;printf("Pleaseinputaarray:

//正确的程序如下：#includevoidmain(){inti,j,n,*tem[3];intp[2][3],q[3][2],r[2][2];tem[0]=p[0];tem[1]=q[0];tem

给,下面是Cholesky分解法的C++经典算法：//-------------------------------------------------------------------//Chol

#includevoidmain(){\x09inta[3][2]={1,2,3,4,5,6};\x09intb[2][3],i,j;\x09for(i=0;i

初始化二维2×3矩阵：DimMyArray(1,2)AsInteger然后运行过程中加入代码定义二维3×2矩阵：ReDimPreserveMyArray(2,1)AsInteger再用一个镶嵌For.

#include#defineN2#defineM3voidmain(void){inti,j;inta[N][M];intb[M][N];for(i=0;i

intfun(intarray[3][3]){inti,j,t;for(i=0;i

#include#include#includevoidmain(){inta[5][5],b[5][5],c[5][5],i,j;srand(time(0));for(i=0;i

//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){inta[3][3]={{1,3,5},{6,7,9},{10,1

#includevoidmain(){intn;printf("请输入一个正整数:");scanf("%d",&n);intsum=0;if(n%2==1){for(inti=1;i

自己敲的,你看看是不是你要的……我吃饭去了,有事发邮件479292539@qq.com#include#defineM4#defineN5voidmain(){inta[M][N],i,j;intnu

intsum=0;for(inti=0;i{if(i%3!=0){sum+=i;}}

#defineM3#defineN3intmain(){inti,j,sum=0,Min,posi=0,posj=0,arry[M][N];for(i=0;i

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

Private Sub Command1_Click()Dim a(4, 2) As Integer, b(2, 4)&

(1)两个矩阵相加后,生成一个新的矩阵.#include<stdio.h>#include<stdlib.h>     &nb

PrivateSubCommand1_Click()RandomizeTimerDimnAsInteger,mAsIntegerDimarr1()AsInteger,arr2()AsIntegerDi

//-------------------------------------//-----------------------------------------------------------

main(){inta[3][3]={{1,2,3},{4,5,6},{7,8,9}}inti,j;intsum=0,s=0;for(i=0;i

publicclassTestArrays{publicstaticvoidmain(String[]args){int[][]a=newint[3][3];int[][]b=newint[3][3]