1、 通过输入对话框输入一批成绩(以输入“-1”表示结束),把及格的和不及格的成绩分别存放在数组a和b中,并以每行5 个数据的形式输出数组a和b。
昨晚弄了好久呢~~~最好玉丰帮忙实现了,但是那代码他自己用了半个小时才弄完的。觉得开始的路就不对。
只好把他那代码珍藏了。“可抄性”太差了~~~































































































题目1、定义一个5×4的二维数组,完成对数组的各元素的输入后.
#include/*定义一个5×4的二维数组,完成对数组的各元素的输入后.要求(1)求各行元素之和,并将这些和按从小到大排列输出.(2)求各列元素之和,并将这些和按从大到小排列输出.(3)对所有元素按
VB 输入一个3行4列的二维数组,分别求出每一行、每一列元素之和 要求
你可以将行列算出的值用一个数组接收,这里是采用变量接收的方式.Option Compare DatabasePrivate Sub aa()Dim h
C语言输入16个数字,输出为4*4的矩形,用二维数组做
对于提问者的补充,我只能这样说:楼上的是用c++做的,cin>>*(C++中用于输入)和cout
c语言编程 将一个随机输入的5*5二维数组旋转后以5行5列输出 旋转方式有4种 分别以列表的对角线旋转
#includevoidprint(intn[5][5]){inti,j;for(i=0;i{for(j=0;j{printf("%d\t",n[i][j]);}printf("\n");}}void
c语言 二维数组 设计一程序,输入一个五行五列的矩阵,计算并显示输出该矩阵四周那一圈元素的合计值
#include #include #define size 5int main(){&
编写一个函数,输入一个整型二维数组,输出其中最大值所在的行号和列号,以及最大值
#include#includevoidmaxp(int**ia,intr,intc,int*x,int*y,int*max){*max=ia[0][0];inti,j;for(i=0;i
C语言:输入一个二维数组(3*4矩阵),输出从左上到右下最佳路径(经过节点的和最小);
答案修改好了,没用函数,只用到了数组之前的知识.(我实在想不出什么好算法了,只好用4层循环嵌套,不知道这样的程序是否合楼主的要求)#include#includeintmain(void){inta[
用二维数组写一个C++程序,要求输入十个整数时可以输出它们的最大值最小值和众数(出现次数最多的数)
能用一维做吗?因为感觉用二维是浪费啊,在说比如这样a【0】【10】是不是不一样相当于一维U盘掉了我有个例题到U盘里不能发给你I0.0!再问:其实我不是很懂。。囧随便用什么吧,能做出来就行,膜拜大神再答
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你设置的是静态数组,静态数组的创建非常方便,使用完也无需释放,要引用也简单,但是创建后无法改变其大小是其致命弱点!就是说a[m][n],m,n不能是变量,你平常可能看到过a[m][n]定义数组的,但仔
如何用C语言编一个程序,输入一个二维数组,然后以一个矩阵的方式输出.
#includevoidmain(){\x09inti,j,a[2][3],*p;\x09for(i=0;i再问:换成一个m*n的矩阵怎么写?再答:m,n是变量吗?数组要确定长度的。
问一道C语言题目用指针编程:输入一个4×5的二维整型数组,输出其中最大值、最小值以及它们所在行和列的下标 #inclu
#includeintmain(void){voidmax(int*);voidmin(int*);intarr[4][5];inti,j;for(i=0;i再问:能帮我看一下我的代码哪里错了吗,谢谢
利用VB,编写一个3*4的二维数组输入任意整数,求所有数组元素和及平均值
Private Sub Command1_Click()Dim i As Integer, j As Integer,&
java 二维数组,输入如下的方阵
我编出来来了,不过5分不足以共享出来,你邮箱我发给你算法思想要是要代码起码要再加10分--!这个代码编了我好久以下是测试用例:请输入矩阵的大小21234请输入矩阵的大小3126357
输入一个3行4列整形二维数组的值,输出最大元素的值及其下标.在主函数中给数组赋值,子函数max中求最大值
下面的程序请参考,并在最后附有运行结果.#include#includevoidmax(inta[][4],intk,int*x,int*y,int*z);intmain(void){inta[3][
c语言编程 将一个随机输入的4*4二维数组逆时针旋转90度后以4行4列输出
#includeintmain(){inta[4][4];for(inti=0;i
用一个循环 输出二维数组
其实二维数组在电脑内存中是连续的.例如:inta[][5]={{1,2,3,4,5},{6,7,8,9,10}};for(inti=0;i
利用随机函数产生16个随机整数给一个4*4的二维数组赋值.按行列输出数组 ,求最外一圈元素之和.
a=round(100*rand(4,4));%生成0-100的随机4*4矩阵a(1,:)%输出第一行...a(:,1)%输出第一列...sum([a(1,:)a(2,1)a(2,4)a(3,1)a(
输入一个4行4列的二维数组,统计偶数的个数,并输出所有小于5的数.
#includeintmain(void){unsignedintdata[4][4];inti,j,odd=0;for(i=0;i
1、 通过输入对话框输入一批成绩(以输入“-1”表示结束),把及格的和不及格的成绩分别存放在数组a和b中,并以每行5 个数据的形式输出数组a和b。
昨晚弄了好久呢~~~最好玉丰帮忙实现了,但是那代码他自己用了半个小时才弄完的。觉得开始的路就不对。
只好把他那代码珍藏了。“可抄性”太差了~~~Dim n1() As Integer
Dim ss As Single
'------------------------------------------------------------
Dim a() As Integer
Dim b() As Integer
Dim C() As Integer
Dim aSize As Integer
Dim bSize As Integer
'------------------------------------------------------------
Private Sub Command1_Click()
aSize = 0
bSize = 0
Dim intA As Integer
Dim intC As Integer
Dim i As Integer
![]()
'初始化
msg$ = "请输入分数(-1结束)"
msgtitle$ = "输入数据"
start:
ss = InputBox(msg$, msgtitle$)
If ss < 0 Or ss > 100 Then
GoTo finish
Else
If ss < 60 Then
If aSize > 0 Then
![]()
ReDim C(UBound(a)) As Integer
For intA = 0 To UBound(a)
C(intA) = a(intA)
Next intA
![]()
End If
![]()
ReDim a(aSize + 1) As Integer
![]()
If aSize > 0 Then
For intC = 0 To UBound(C)
a(intC) = C(intC)
Next intC
a(aSize + 1) = ss
End If
a(aSize) = ss
aSize = aSize + 1
End If
GoTo start
End If
finish:
For i = 0 To UBound(a)
Print a(i)
Next i
![]()
End Sub
Private Sub Form_Load()
End Sub
转载于:https://www.cnblogs.com/gengxiaochao/archive/2008/01/15/1039627.html
VB实验报告“找出二维数组n×m中的鞍点”
姓名:叶大塽
班级及学号:电气2班15050342035
日期:2016年5月26日
一.实验目的
设计Vb程序,找出二维数组n×m中的鞍点。
二.实验内容
找出二维数组n×m中的鞍点,所谓的鞍点是指它在本行中值最大,在本列中值最小,输出鞍点的行,列,有可能在一个数组中找不到一个鞍点,如无鞍点则输出“无”。
三.实验程序
Private Sub Command1_Click()
Dim i As Integer, j As Integer, c As Integer, r As Integer
Dim n As Integer
cs1
xs
For i = 1 To 4
For j = 1 To 4 Private Sub Command1_Click()
Dim i As Integer, j As Integer, c As Integer, r As Integer
Dim n As Integer
cs1
xs
For i = 1 To 4
For j = 1 To 4
If a(i, 0) < a(i, j) Then
c = j
a(i, 0) = a(i, j)
End If
Next
n = 0
For j = 1 To 4
If a(i, 0) < a(j, c) Then
Print "+";
n = n + 1
End If
Next
If n = 3 Then Print "a(" & i & "," & c & ")"; a(i, c)
Next
xs
End Sub
Public a(4, 4) As Integer
Public Sub xs()
Dim i As Integer, j As Integer
For i = 1 To 4
For j = 0 To 4
Form1.Print a(i, j); " ";
Next
Form1.Print
Next
End Sub
Public Sub cs0()
a(1, 1) = 1
a(1, 2) = 1
a(1, 3) = 1
a(1, 4) = 1
a(2, 1) = 1
a(2, 2) = 1
a(2, 3) = 1
a(2, 4) = 1
a(3, 1) = 1
a(3, 2) = 1
a(3, 3) = 1
a(3, 4) = 1
a(4, 1) = 1
a(4, 2) = 1
a(4, 3) = 1
a(4, 4) = 1
End Sub
Public Sub cs1()
a(1, 1) = 1
a(1, 2) = 2
a(1, 3) = 3
a(1, 4) = 4
a(2, 1) = 1
a(2, 2) = 1
a(2, 3) = 1
a(2, 4) = 5
a(3, 1) = 1
a(3, 2) = 1
a(3, 3) = 1
a(3, 4) = 6
a(4, 1) = 1
a(4, 2) = 1
a(4, 3) = 1
a(4, 4) = 7
End Sub
Public Sub cs2()
a(1, 1) = 1
a(1, 2) = 1
a(1, 3) = 5
a(1, 4) = 1
a(2, 1) = 1
a(2, 2) = 2
a(2, 3) = 4
a(2, 4) = 3
a(3, 1) = 1
a(3, 2) = 1
a(3, 3) = 6
a(3, 4) = 1
a(4, 1) = 1
a(4, 2) = 1
a(4, 3) = 7
a(4, 4) = 1
End Sub
Public Sub cs3()
a(1, 1) = 1
a(1, 2) = 5
a(1, 3) = 1
a(1, 4) = 1
a(2, 1) = 1
a(2, 2) = 4
a(2, 3) = 2
a(2, 4) = 3
a(3, 1) = 1
a(3, 2) = 6
a(3, 3) = 1
a(3, 4) = 1
a(4, 1) = 1
a(4, 2) = 7
a(4, 3) = 1
a(4, 4) = 1
End Sub
If a(i, 0) < a(i, j) Then
c = j
a(i, 0) = a(i, j)
End If
Next
n = 0
For j = 1 To 4
If a(i, 0) < a(j, c) Then
Print "+";
n = n + 1
End If
Next
If n = 3 Then Print "a(" & i & "," & c & ")"; a(i, c)
Next
xs
End Sub
Public a(4, 4) As Integer
Public Sub xs()
Dim i As Integer, j As Integer
For i = 1 To 4
For j = 0 To 4
Form1.Print a(i, j); " ";
Next
Form1.Print
Next
End Sub
Public Sub cs0()
a(1, 1) = 1
a(1, 2) = 1
a(1, 3) = 1
a(1, 4) = 1
a(2, 1) = 1
a(2, 2) = 1
a(2, 3) = 1
a(2, 4) = 1
a(3, 1) = 1
a(3, 2) = 1
a(3, 3) = 1
a(3, 4) = 1
a(4, 1) = 1
a(4, 2) = 1
a(4, 3) = 1
a(4, 4) = 1
End Sub
Public Sub cs1()
a(1, 1) = 1
a(1, 2) = 2
a(1, 3) = 3
a(1, 4) = 4
a(2, 1) = 1
a(2, 2) = 1
a(2, 3) = 1
a(2, 4) = 5
a(3, 1) = 1
a(3, 2) = 1
a(3, 3) = 1
a(3, 4) = 6
a(4, 1) = 1
a(4, 2) = 1
a(4, 3) = 1
a(4, 4) = 7
End Sub
Public Sub cs2()
a(1, 1) = 1
a(1, 2) = 1
a(1, 3) = 5
a(1, 4) = 1
a(2, 1) = 1
a(2, 2) = 2
a(2, 3) = 4
a(2, 4) = 3
a(3, 1) = 1
a(3, 2) = 1
a(3, 3) = 6
a(3, 4) = 1
a(4, 1) = 1
a(4, 2) = 1
a(4, 3) = 7
a(4, 4) = 1
End Sub
Public Sub cs3()
a(1, 1) = 1
a(1, 2) = 5
a(1, 3) = 1
a(1, 4) = 1
a(2, 1) = 1
a(2, 2) = 4
a(2, 3) = 2
a(2, 4) = 3
a(3, 1) = 1
a(3, 2) = 6
a(3, 3) = 1
a(3, 4) = 1
a(4, 1) = 1
a(4, 2) = 7
a(4, 3) = 1
a(4, 4) = 1
End Sub
四.实验问题及解决方法
对程序无从下手,还的继续学习。Vb的确是很实用的软件,学好前途无量。但是也确实难学,学了这么久了,我还没有看出其中的要点,还不能真正入门,感觉太难。以后再接再厉吧
总时间限制:1000ms内存限制:65536kB描述
输入第一行包含两个整数n和m,表示矩阵A的行数和列数。1 <= n <= 100,1 <= m <= 100。
接下来n行,每行m个整数,表示矩阵A的元素。相邻两个整数之间用单个空格隔开,每个元素均在1~1000之间。输出m行,每行n个整数,为矩阵A的转置。相邻两个整数之间用单个空格隔开。样例输入
3 3 1 2 3 4 5 6 7 8 9
样例输出
1 4 7 2 5 8 3 6 9
源代码(一)
#include <iostream> #include <algorithm> using namespace std; int main() { int n,m; int a[101][101]; int b[101][101]; cin>>n>>m; for(int i=0; i<n; i++) for(int j=0; j<m; j++) { cin>>a[i][j]; } for(int i=0; i<n; i++) for(int j=0; j<m; j++) { b[j][i]=a[i][j]; } for(int i=0; i<m; i++) { for(int j=0; j<n; j++) { cout<<b[i][j]<<" "; } cout<<endl; } return 0; }
源代码(二)
#include <iostream> #include <algorithm> using namespace std; int main() { int A[101][101], n, m, i, j; cin >> n >> m; for(i=0;i<n;i++) for(j=0;j<m;j++) cin >> A[i][j]; for(i=0;i<m;i++){ for(j=0;j<n;j++){ cout << A[j][i] << " "; } cout << endl; } }