从1位置起始，到头就改变方向，直至数字达到nm停止 ``` #include <bits/stdc++.h> using namespace std; const int N = 110; int mp[N][N]; int main() { int n, m; scanf("%d %d", &n, &m); int posx = 1, posy = m; int num = 1; int direction = 1; while (num <= n * m) { mp[posx][posy] = num++; if (direction == 1) { if (posy < m && !mp[posx][posy + 1]) posy++; else { direction = 2; posx++; } } else if (direction == 2) { if (posx < n && !mp[posx + 1][posy]) posx++; else { direction = 3; posy--; } } else if (direction == 3) { if (posy > 1 && !mp[posx][posy - 1]) posy--; else { direction = 4; posx--; } } else { if (posx > 1 && !mp[posx - 1][posy]) posx--; else { direction = 1; posy++; } } } for (int i = 1; i <= n; ++i) { for (int j = 1; j <= m; ++j) { printf("%5d", mp[i][j]); } printf("\n"); } return 0; } ```

找固定移动模式，模拟即可 ```cpp #include <bits/stdc++.h> #define int long long using namespace std; const int N = 110; int n,m,a[N][N],b[N][N]; void solve(){ cin>>m>>n; int x=0,y=n-1,cnt=1,d=0; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); b[0][n-1] = 1; a[0][n-1] = 1; while(cnt<n*m){ while(x<m-1-d&&!a[x+1][y]){ b[++x][y] = ++cnt; a[x][y] = 1; }while(y!=d&&!a[x][y-1]){ b[x][--y] = ++cnt; a[x][y] = 1; }while(x!=d&&!a[x-1][y]){ b[--x][y] = ++cnt; a[x][y] = 1; } while(y<n-1&&!a[x][y+1]){ b[x][++y] = ++cnt; a[x][y] = 1; } d+=1; } for(int i=0;i<m;i++){ for(int j=0;j<n;j++){ printf("%5lld",b[i][j]); }printf("\n"); } } signed main() { ios::sync_with_stdio(0); cin.tie(0); int t=1; while(t--)solve(); return 0; } ```

``` #include<iostream> #include<vector> #include<iomanip> using namespace std; int main() { int m, n; int num = 1; int flag = 1; cin >> m >> n; int x = 0, y = n - 1; vector<vector<int>> arr(m, vector<int>(n, 0)); for(int i = 0; i < m * n; i++) { arr[x][y] = num++; if(flag == 1) { if(x < m - 1 && arr[x + 1][y] == 0) { x++; } else { flag = 2; y--; } } else if(flag == 2) { if(y > 0 && arr[x][y - 1] == 0) { y--; } else { flag = 3; x--; } } else if(flag == 3 ) { if(x > 0 && arr[x - 1][y] == 0) { x--; } else { flag = 4; y++; } } else if(flag == 4) { if(y < n - 1 && arr[x][y + 1] == 0) { y++; } else { flag = 1; x++; } } } for(int i = 0; i < m; i++) { for(int j = 0; j < n; j++) { printf("%5d", arr[i][j]); } cout << endl; } return 0; } ```

include include include using namespace std; int a[110][110]; int main() { int tot, x, y, m, n; cin >> m >> n; memset(a, 0, sizeof(a)); x = 0; y = n - 1; a[x][y] = 1; tot = 1; while (tot = 0 && !a[x][y - 1]) a[x][--y] = ++tot; while (x - 1 >= 0 && !a[x - 1][y]) a[--x][y] = ++tot; while (y + 1 < n && !a[x][y + 1]) a[x][++y] = ++tot; } for (int i = 0; i < m; i++) { for (int j= 0; j < n; j++) { printf("%5d", a[i][j]); } cout << endl; } cout << endl; return 0; }

``` #include <iostream> #include <cstring> #include <algorithm> using namespace std; const int N = 110; int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1}; int g[N][N]; int n, m; int main() { cin >> n >> m; int d = 2, x = 0, y = m - 1; for (int i = 1; i <= n * m; i ++ ) { g[x][y] = i; int a = x + dx[d], b = y + dy[d]; if(a < 0 || a >= n || b < 0 || b >= m || g[a][b]) { d = (d + 1) % 4; a = x + dx[d], b = y + dy[d]; } x = a, y = b; } for (int i = 0; i < n; i ++ ) { for (int j = 0; j < m; j ++ ) printf("%5d", g[i][j]); cout << endl; } return 0; } ```

重点理解while循环的内容，理解蛇形填数的排布 ``` #include<bits/stdc++.h> using namespace std; const int N=110; int a[N][N]; int m,n; //矩阵行、列 int x,y; //数字 1 的坐标 int number; // 记录数字的个数 int main(){ cin>>m>>n; memset(a,0,sizeof(a)); //初始化 1 的坐标 (0,n-1) x=0; y=n-1; a[x][y]=1; number=1; while(number<m*n){ //顺时针填写数字 while(x+1 < m && !a[x+1][y]) a[++x][y] = ++number; //从上到下：x轴++，y轴不变 ：x++,number++,a[x][y]=number; while(y-1 >= 0 && !a[x][y-1]) a[x][--y] = ++number; //从右到左：x轴不变，y轴-- while(x-1>=0 && !a[x-1][y]) a[--x][y] = ++number; //从下到上 ：x轴--，y轴不变 while(y+1<n && !a[x][y+1]) a[x][++y] = ++number; //从左到右 ：x轴不变，y轴++ } //打印出来 for(int i=0;i<m;i++){ for(int j=0;j<n;j++) printf("%5d",a[i][j]); cout<<endl; } cout<<endl; return 0; } ```

``` #include <cstdio> #include <cmath> #include<iostream> using namespace std; int a[200][200]; int flag[200][200]; int main(){ int m,n; cin>>m>>n; int x=n,y=1; int count=1; int num=1; while(num<=n*m){ while(y<=m&&flag[x][y]!=1&&num<=n*m){ flag[x][y]=1; a[x][y]=num; num++; y++; } x--; y--; while(x>=1&&flag[x][y]!=1&&num<=n*m){ flag[x][y]=1; a[x][y]=num; num++; x--; } x++; y--; while(y>=1&&flag[x][y]!=1&&num<=n*m){ flag[x][y]=1; a[x][y]=num; num++; y--; } y++; x++; while(x<=n&&flag[x][y]!=1&&num<=n*m){ flag[x][y]=1; a[x][y]=num; num++; x++; } x--; y++; } for(int i=1;i<=m;++i){ for(int j=1;j<=n;++j){ printf("%5d",a[j][i]); } cout<<endl; } return 0; } ```

``` // #include <iostream> #include <cstdio> using namespace std; int res[100][100]; int main() { int m,n; cin>>m>>n; int dx[]={0,1,0,-1},dy[]={1,0,-1,0}; //偏移量 for(int x=0,y=n-1,d=0,k=1; x<m,y<n,k<=m*n ;k++) { res[x][y]=k; int a=x+dx[d],b=y+dy[d]; if(a<0||a>=m||b>=n||b<0||res[a][b]) //res[a][b]是重复格子的情况，初始化数组为零，如果走过一遍格子会有数字，需要转弯 { d=(d+1)%4; a=x+dx[d],b=y+dy[d]; }//看是否撞墙或者走到重复的格子 x=a,y=b;//没有的话就更新位置 } for(int i=0;i<m;i++) { for(int j=0;j<n;j++) { printf("%5d",res[i][j]); } cout<<endl; } return 0; } ```

include using namespace std; typedef long long ll; define endl '\n' int main() { int m,n; scanf("%d %d",&m,&n); int dir=1; vector > arr(m,vector (n)); // int arr[100][100]; int x,y,bound_x,bound_y; bound_x=m; bound_y=n; x = 0; y = n-1; int cir = 0; for(int cnt=1;cnt<=mn;cnt++) { ``` if(dir==1) { arr[x][y] = cnt; x++; if(x == (bound_x-cir)) { dir = 2; x--; y--; } } else if(dir==2) { arr[x][y] = cnt; y--; if(y == (cir-1)) { dir = 3; y++; x--; } } else if(dir==3) { arr[x][y] = cnt; x--; if(x == (cir -1)) { dir =4; x++; y++; } } else if(dir ==4 ) { arr[x][y] = cnt; y++; if(y == (n-cir-1)) { dir = 1; y--;x++; cir++; } } } for(int i = 0; i < m;i++) { for(int j = 0;j < n;j++) { printf("%5d",arr[i][j]); } cout<<endl; } return 0; ``` }

``` #include <bits/stdc++.h> using namespace std; const int N = 1010; int a[N][N]; int main() { int m, n; cin >> m >> n; //初始位置为右上角 int y = n; int x = 1; int cnt = 1; while (cnt <= m \* n) { // 下 while (x <= m && a[x][y] == 0) { a[x][y] = cnt++; x++; } x--;//这里x--是因为最后下走的时候x多加了一次 下面同理 y--; // 左 while (y >= 1 && a[x][y] == 0) { a[x][y] = cnt++; y--; } y++; x--; // 上 while (x >= 1 && a[x][y] == 0) { a[x][y] = cnt++; x--; } x++; y++; // 右 while (y <= n && a[x][y] == 0) { a[x][y] = cnt++; y++; } y--; x++; } // 输出 for(int i = 1;i <= m;i++) { for(int j = 1;j <= n;j++) { cout << setw(5) << a[i][j]; } cout << '\\n'; } return 0; } ```