我该如何打印 - [英] How do I print this-
问题描述
______ 4
____4 3 4
__4 3 2 3 4
4 3 2 1 2 3 4
__4 3 2 3 4
___ 4 3 4
______4
写一个C ++程序来获取双金字塔上方。
'_' - 这表示空间，就是金字塔。
我尝试过：
#include< iostream>使用namespace std
;
int main（）
{
int i，j;
for（i = 4; i> = 1; i - ）{
for（j = 1; j< = i; j ++）{
cout<<;
}
for（int j = 4; j> = i; j - ）{
cout<< j;
}
for（j = i + 1; j< = 4; j ++）{
cout<< j;
cout<<'\ n';
}
for（i = 2; i< = 4; i ++）{
for（j = 1; j< i; j ++）{
cout<< j;
}
for（j = 1 + i; j< = 4; j ++）{
cout<< j;
cout<<'\ n';
}
}
}
返回0;
}
我试过这个，我觉得我做了一些非常愚蠢的错误。请帮助我。
首先正确地缩进你的代码，或者，如果你打算使用可执行的1TB，那么至少要始终如一：
int main（）{
int i，j;
for（i = 4; i> = 1; i - ）{
for（j = 1; j< = i; j ++）{
cout<<;
}
for（int j = 4; j> = i; j - ）{
cout<< j;
}
for（j = i + 1; j< = 4; j ++）{
cout<< j;
cout<<'\ n';
}
for（i = 2; i< = 4; i ++）{
for（j = 1; j< i; j ++）{
cout<< j;
}
for（j = 1 + i; j< = 4; j ++）{
cout<< j;
cout<<'\ n';
}
}
}
返回0;
}现在你可以看到你在做什么。我想到的第一件事就是：你有两个循环，都改变
i，其中一个在另一个里面...所以你的代码永远不会退出。 / blockquote>
让我们猜测输出有太多行
for （ j = i + 1; j< = 4 ; j ++）{
cout<< j;
cout<< ' \ n'; // 删除此行
}
cout<< ' \ n'; // 并将其放在那里
你的程序很复杂，只有4号。
a小分析应该有助于让它变得更好。钻石具有垂直和水平对称性，全部围绕中心（1）
尺寸行Cols Center
1 1 1 0 ， 0
2 3 3 1 ， 1
3 5 5 2 ， 2
4 7 7 3 ， 3
将大小存储在变量中并减少行，列和中心应该不复杂。
你能够 请注意，所有'2'都距离中心1 col或1行。
'3'更进一步：行距离+ cols距离给出位置值。
如果值更大，请打印一个空格。
我让你填补这些洞作为练习。
建议：
- 学习一种或多种分析方法，EW Djikstra / N. Wirth Stepwize Refinment / top-Down方法是一个良好的开端。
Structured Programming.pdf [ ^ ]
https：//en.wikipedia。 org / wiki / Top-down_and_bottom-up_design [ ^ ]
https://en.wikipedia.org/wiki/Structured_programming [ ^ ]
https://en.wikipedia.org/wiki/Edsger_W._Dijkstra [ ^ ]
https://www.cs.utexas.edu/users/EWD/ewd03xx/EWD316.PDF [ ^ ]
______4
____4 3 4
__4 3 2 3 4
4 3 2 1 2 3 4
__4 3 2 3 4
___ 4 3 4
______4
Write a C++ program to obtain the above double pyramid.
'_' - this indicates space, that's how it is a pyramid.
What I have tried:
#include <iostream> using namespace std; int main() { int i,j; for(i=4; i>=1; i--){ for(j=1; j<=i; j++){ cout<<" "; } for(int j=4; j>=i; j--){ cout<<j; } for( j=i+1; j<=4; j++){ cout<<j; cout<<'\n'; } for(i=2; i<=4; i++){ for(j=1; j<i; j++){ cout<<j; } for(j=1+i; j<=4; j++){ cout<<j; cout<<'\n'; } } } return 0; }
I have tried this and I have a feeling that I have done some really stupid mistake. Please help me.解决方案Start by indenting your code properly, or, if you are going to use the execrable 1TB then at least consistently:
int main(){ int i,j; for(i=4; i>=1; i--){ for(j=1; j<=i; j++){ cout<<" "; } for(int j=4; j>=i; j--){ cout<<j; } for( j=i+1; j<=4; j++){ cout<<j; cout<<'\n'; } for(i=2; i<=4; i++){ for(j=1; j<i; j++){ cout<<j; } for(j=1+i; j<=4; j++){ cout<<j; cout<<'\n'; } } } return 0; }And now you can see what you are doing. And the first thing that springs to mind is this: you have two loops, both altering
iand one of them is inside the other ... so your code never exits.
Lets guess the output have too many lines
for( j=i+1; j<=4; j++){ cout<<j; cout<<'\n'; // Remove this line } cout<<'\n'; // and put it there
Your program is complicated and for size 4 only.
a little analyze should help to get it better. The diamond have a vertical and horizontal symmetry, all is around the center (1)
Size Rows Cols Center 1 1 1 0,0 2 3 3 1,1 3 5 5 2,2 4 7 7 3,3
Should not be complicated to store the Size in a variable and reduce Rows, Cols and Center.
You can notice that all '2' are at a distance of 1 col or 1 row from center.
'3' are 1 step further: distance in rows + distance in cols give you the value at position.
if value is more the the size, print a space.
I let you fill the holes as an exercise.
Advice:
- Learn one or more analyze methods, E.W. Djikstra/N. Wirth Stepwize Refinment/top-Down method is a good start.
Structured Programming.pdf[^]
https://en.wikipedia.org/wiki/Top-down_and_bottom-up_design[^]
https://en.wikipedia.org/wiki/Structured_programming[^]
https://en.wikipedia.org/wiki/Edsger_W._Dijkstra[^]
https://www.cs.utexas.edu/users/EWD/ewd03xx/EWD316.PDF[^]
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