不规则的间距书写呼吸道模式
#include<iostream>
using namespace std;
int main(){
int n;
cin>>n;
for(int i=1;i<=n;i++){
int space=n-i;
for(int j=1;j<=space;j++){
cout<<" ";
}
for(int k=i;k>=1;k--){
cout<<k<<" ";
}
for(int l=2;l<=i;l++){
cout<<l<<" ";
}
cout<<endl;
}
return 0;
}
我写了此代码以打印引导模式，但是我收到的输出具有不规则的间距，无论我在代码中找不到错误。
#include<iostream>
using namespace std;
int main(){
int n;
cin>>n;
for(int i=1;i<=n;i++){
int space=n-i;
for(int j=1;j<=space;j++){
cout<<" ";
}
for(int k=i;k>=1;k--){
cout<<k<<" ";
}
for(int l=2;l<=i;l++){
cout<<l<<" ";
}
cout<<endl;
}
return 0;
}
I wrote this code for printing a palindrome pattern but the output I received has irregular spacing and no matter what I couldn't find the error in the code.
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评论（1）
我假设“正确”的间距是 其中所有
1的阵容 。您可以尝试以下一些步骤来解决问题：1。
1的位置如何确定？该位置应由最后一行确定，从
n开始 - 因为根据您的程序，我们不需要前面的任何空间。2。
1在哪个位置出现在最后一行？在字符
1之前，有空格和其他数字 - n-1 数字从n到2，并且同样是相同数量的空间，因为每个字符之后都会显示1个空间，除了最后一个空间，在这里没关系。总共 2（n-1）字符1。3。最后，在行
k？之前需要多少个空间
对于行
k，我们可以使用上面的步骤知道字符1有 2（k-1）字符，但是我们需要 2（N-1）填充长度。因此，我们可以以 2（n-1）-2（k-1）= 2（nk）来获得线路前的所需空间。这就是为什么您需要在第7行上的“ Times 2”。尝试找到问题的步骤。
I assume the "correct" spacing is where all the
1's line up. Here are some steps you could try to solve the problem:1. How is the position of
1determined?The position should be determined by the last line, starting with
n— since we don't require any spaces on the front, according to your program.2. At which position will
1appear on the last line?Before the character
1, there are spaces and other numbers — n-1 numbers fromnto2, and also the same number of spaces, since 1 space is presented after every character, except the last, which doesn't matter here. In total, 2(n-1) characters before1.3. Finally, How many spaces are required before line
k?For line
k, we can use the above step to know that before the character1there are 2(k-1) characters, but we need 2(n-1) to fill the length. Thus, we can obtain the required space in front of the line as 2(n-1)-2(k-1) = 2(n-k). That's why you need the "times 2" over there on line 7.Try to find steps toward problems.