10 条题解

  • 2
    @ 2024-4-8 16:42:44

    从1位置起始，到头就改变方向，直至数字达到n*m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;
}
    • 1
      @ 2025-4-3 9:19:00

      找固定移动模式，模拟即可

      #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;
}
      • 0
        @ 2025-3-25 16:36:00 
        #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;
}
        • 0
          @ 2025-3-18 21:48:31

          #include <iostream> #include <cstring> #include <cstdio>

          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 < m * n) {
    while (x + 1 < m && !a[x + 1][y]) a[++x][y] = ++tot;
    while (y - 1 >= 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;

          }

          • 0
            @ 2024-4-18 20:09:49 
            #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;
}
            • 0
              @ 2024-4-14 21:33:25 
              #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;
}
              • 0
                @ 2024-4-12 17:07:28 
                //
#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;
}
                • 0
                  @ 2024-4-11 20:39:23

                  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;
}
                  • 0
                    @ 2024-4-11 14:22:17

                    #include<bits/stdc++.h> using namespace std; typedef long long ll; #define endl '\n'

                    int main() { int m,n; scanf("%d %d",&m,&n); int dir=1; vector<vector<int>> arr(m,vector<int>(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<=m*n;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;

                    }

                    • 0
                      @ 2024-4-10 11:52:14 
                      #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;

}
                      • 1

                      蛇形填数2

                      信息

                      ID
                      78
                      时间
                      1000ms
                      内存
                      256MiB
                      难度
                      7
                      标签
                      递交数
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                      已通过
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