用C ++打印模式 [英] Printing a pattern in C++

问题描述

模式如下：



1. _ _ _ *

2. _ _ * *

3. _ * * *

4. * * * *



[忽略数字：只是用它们来增加可读性，和下划线是这里的空间用于表示。]



很好用2 for或while循环很容易做到这一点，但我的朋友只用一个循环挑战我。我必须为它创建一个通用程序，这样它不仅可以运行4个，也可以运行5个或6个。



我试过试过，直到现在我已达到这个目的：

The pattern is as follows:

1. _ _ _ *
2. _ _ * *
3. _ * * *
4. * * * *

[Ignore numbers: just used them to increase readiblity, and underscores are spaces here for representation.]

Well its quiet easy to do it with 2 for or while loops , but my friend challenged me to it with only 1 loop.And i have to create a general program for it , so that it runs for not only 4 but 5 or 6 too.

Well i tried and tried , and till now i have reached this far:

int a=1;
int times=4; // no. of times a line , 4 here but user can enter anything
int totalelements=times*times;
int n=1;
int breakpoint=(times*n)-(n-1);

// breakpoints are the first occurences of * like 4th position in 1st line, 7th in 2nd line..


while(a<(totalelements+1))

{

if (breakpoint==a)
{cout<<"*"; 
n++;                            //to increment n for next value in breakpoint formula
breakpoint=(times*n)-(n-1); }   //get new breakpoint formula

else cout<<" ";

if(a%times==0) { cout<<endl;} // for printing nextline every 4th element here

}

预期产量为：



1. _ _ _ *

2. _ _ * _

3. _ * _ _

4. * _ _ _



所以我做错的事情是：我能够找到它的断点模式，但它只在那些断点打印星星，而不是在那之后。 />
我应该怎么做，在那个断点之后打印那些星星



我能说的一件事就是不行。正在打印的星星是当时n的值，所以我应该如何链接它？



程序必须只使用1个循环，否则不能。变量，没有goto语句。

The expected output is :

1. _ _ _ *
2. _ _ * _
3. _ * _ _
4. * _ _ _

So the thing i am doing wrong is : i am able to find out breakpoint pattern for it , but it prints stars only at those breakpoints , not after that.
How should i do that , to print those stars after that breakpoint

one thing i can relate is the no. of stars being printed is the value of n at that time, so how should i link it?

The program must use only 1 loop , any no. of variables , and no goto statement .

推荐答案

我认为你可以简化问题，只需要识别所需模式的二维性质，即你有行 列：

I think you may simplify the matter just recognizing the bidimensional nature of the required pattern, that is you have rows an columns:

#include <iostream>
using namespace std;
int main()
{
    int n=4; // no. of times a line , 4 here but user can enter (almost :-) ) anything 
    for (int i=0; i<n*n; ++i)
    {

         int r = i/n; // row number
         int c = i%n; // column number

         cout << (c < n-1-r ? ' ':'*');
         if (c==n-1) cout << endl;
    }
}

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