86 lines
2.4 KiB
Markdown
86 lines
2.4 KiB
Markdown
# $C++$语法基础课$20230212$讲义
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## [$AcWing$ $753$. 平方矩阵 $I$](https://www.acwing.com/problem/content/755/)
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<center><img src='http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/02/d2c3c1478f9b46e9ad622e337f0c9292.png' width="480" height="360"></center>
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**题解：**
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1. 这是一道找规律的题，一般都是思考行与列的数据关系
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2. 从左侧看，是找出行与列的最小值
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3. 从右侧看，上面的办法似乎行不通，可以变通一下，假设我们是从右向左看，就是$n-i+1$就可以了~
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```cpp {.line-numbers}
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#include<iostream>
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using namespace std;
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int n;
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int main(){
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while(cin>>n,n){
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for(int i=1;i<=n;i++){
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for(int j=1;j<=n;j++)
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printf("%d ",min(min(i,j),min(n+1-i,n+1-j)));
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puts("");
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}
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puts("");
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}
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return 0;
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}
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```
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## [$AcWing$ $754$. 平方矩阵 $II$](https://www.acwing.com/problem/content/description/756/)
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<center><img src='https://cdn.acwing.com/media/article/image/2021/11/05/114892_ec7836a13d-111.png'></center>
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**题解：**
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1. 这是一道找规律的题，类似平方矩阵I，这类题大多通过找横纵坐标$(i/j)$的关系，得到规律
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2. 通过观察发现，对角线$i、j$相等之处，是$1$.其他的数是由$1$往两边发散
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3. 继续通过观察，判断(除$i、j$相等之处)的数，都是横纵坐标之差的绝对值$+1$得到，
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```cpp {.line-numbers}
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#include<iostream>
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using namespace std;
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int main(){
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int n;
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while(cin>>n && n){
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for(int i=1;i<=n;i++){
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for(int j=1;j<=n;j++)
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printf("%d ",abs(i-j)+1);
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puts("");
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}
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puts("");
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}
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return 0;
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}
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```
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## [$AcWing$ $755$. 平方矩阵 III](https://www.acwing.com/problem/content/description/757/)
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<center><img src='http://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/BlogImages/2023/02/f3055756d4bd5c904855d01e7b3e5715.png' width="480" height="360"></center>
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**题解**：
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1. 需要知道$2^n$在$C++$中的计算办法`n<<1`
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2. 其它，就没有了
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```cpp {.line-numbers}
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#include <iostream>
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using namespace std;
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int n;
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const int N=110;
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int main()
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{
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while(cin>>n , n){
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for(int i=0;i<n;i++){
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for(int j=0;j<n;j++)
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printf("%d ",1<<(i+j));
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puts("");
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}
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puts("");
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}
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return 0;
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}
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``` |