#include<bits/stdc++.h> using namespace std; int a[1001][1001]; int main(){ int n; cin>>n; int cnt=1; for(int i=1;i<=n;i++) { if(i%2==1) { for(int j=1;j<=n;j++) { a[i][j]=cnt++; } } else for(int j=n;j>=1;j--) a[i][j]=cnt++; } for (int i =1; i <=n; ++i) { for (int j =1; j <=n; ++j) { cout<<a[i][j]<<" "; } cout<<endl; } return 0; }

这道题的难点在于如何让数字反过来放在二维数组里，具体见下 看完题解，加入团队

#include <bits/stdc++.h> using namespace std; int a[1001][1001]; int main() { int n; cin>>n; int cnt=1; for(int i=1;i<=n;i++) { if(i%4==2 ) for(int j=1;j<=n;j++) a[i][j]=cnt++; else for(int j=n;j>=1;j--) a[i][j]=cnt++; } return 0; }

#include <iostream> using namespace std; int main() { int a,b[1000000]; cin>>a; for(int i=1;i<=a;i++){ b[i]=i; }

→ → → ↓ ↓ ← ← ← → → → ↓ ↓ ← ← ←

答案： #include <bits/stdc++.h> using namespace std; int main(){ int n; cin>>n; for(int i=0;i<n;i++){ if(i%20){ for(int j=1;j<=n;j++){ cout<<j+n*i<<' '; } }else if(i%21){ for(int j=n;j>=1;j--){ cout<<j+n*i<<' '; } } cout<<endl; } return 0; }

#include<iostream> using namespace std; int main(){ int n; cin>>n; for(int i=1;i<=n;i++){ if(i%2!=0){ for(int l=ni-n+1;l<=ni;l++){ cout<<l<<" "; } cout<<endl; }else{ for(int l=ni;l>=ni-n+1;l--){ cout<<l<<" "; } cout<<endl; } } return 0; }

#include <bits/stdc++.h> using namespace std; int main() { int n; scanf("%d", &n); int list[n][n]; int a = 1, b = 1; int count = 1; for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { list[i][j] = 0; } } for (int i = 0; i < n; ++i) { if (a) { for (int j = 0; j < n; ++j) { list[i][j] = b; ++b; } a = 0; } else { for (int j = n - 1; j >= 0; --j) { list[i][j] = b; ++b; } a = 1; } } for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { printf("%d ", list[i][j]); } printf("\n"); } return 0; }