编程题 输入一个5*6矩阵的值,求其元素的和并输入
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#include<stdio.h>#define ROW 3//行数#define COL 4//列数void PlusMatrix(in
用二维数组,再加上必要的逻辑判断对输入矩阵进行转换即可,最后是每项输出对应一个函数,分布处理.
#includeintmain(){inta[5][5]={{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5}};intsum=0,
#includeintmain(){inti;voidfun(inta);printf("输入数字:");scanf("%d",&i);fun(i);return0;}voidfun(inta){if
#include"stdio.h"intmain(){ inta[5][5]={0};
if(j==4)\x09\x09\x09\x09printf("%d\n",max);去掉if(j==4)加大括号.改成这样:#defineM3#defineN5#includevoidmain(){
#includeusingnamespacestd;voidmain(){intCArray[5][5];inti,j;intsum=0;cout
//输入范例假设n赋值为4/*1234(回车)2345(回车)3456(回车)5678(回车)注意输入时数字间要有空格间隔*/#includevoidmain(){inti,j;intsum=0;//
PrivateSubCommand1_Click()DimaAsLongDimiAsLongDimsAsBooleans=Truea=InputBox("请输入一个自然数","输入")'-------
#include#defineN5voidmain(){\x09intmatrix[N][N];\x09intsum=0;\x09inti,j;\x09printf("请输入一个%d*%d的矩阵:\n
#include<stdio.h>void main(){\x05char n1,n2,n3,max;\x05scanf("%c%c%c",&
PrivateSubCommandButton1_Click()DimnNumAsLongIfIsNumeric(TextBox1.Text)ThennNum=Val(TextBox1.Text)If
classArray{public:inta[4][4];Array();//无参构造函数,将矩阵各元素都设为0voidinput(int*);Arrayoperator+(Array,Array)}
你这个是转置吗?先输入一个矩阵Am*n阶的,转置里面应先初始化一个矩阵Bn*m的记得要先初始化哦!然后利用for循环两层i,jfori=0:n-1forj=0:m-1B(i,j)=B(j,i);然后就
#includeintmain(void){inti,j,sum=0;inta[3][3];printf("Inputthearray:\n");for(i=0;i
#includemain(){inta[3][3];inti,j,sum=0;for(i=0;i
使用这段程序就可以实现了a=magic(4);A={};n1=nchoosek([1234],1);fori=1:size(n1,1)b=a(:,n1(i));%1列A=[Ab];endn2=ncho
#include<stdio.h>int main(){ int n,i,j,k=1;
#include <stdio.h>void main(){\x05int i, j, array[5][5], sum=0;\
#includeintmain(){inti=0;printf("pleaseputanumber:\n->");scanf("%d",&i);while(i){printf("%d\n