2983. 蛇行图案

Saitama

重写基本题目

#include <bits/stdc++.h>

using namespace std;

void solve()
{
    int n;
    cin >> n;
    int a[101][101];
    memset(a, 0, sizeof(a));
    int c = 1, x = 0, y = 0;
    while(c < n * n)
    {
        while(x < n - 1 && a[y][x + 1] == 0)
            a[y][x++] = c++;
        while(y < n - 1 && a[y + 1][x] == 0)
            a[y++][x] = c++;
        while(x > 0 && a[y][x - 1] == 0)
            a[y][x--] = c++;
        while(y > 0 && a[y - 1][x] == 0)
            a[y--][x] = c++;
    }
    a[y][x] = c;
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
            cout << a[i][j] << ' ';
        cout << endl;
    }
    return;
}

int main()
{
    int T;
    cin >> T;
    for(int i = 0; i < T; i++)
    {
        printf("case #%d:\n", i);
        solve();
    }
    return 0;
}
10185101178

c期中考试题

include

include

int main()
{
int x; /二维数组的第一个下标/
int y; /二维数组的第二个下标/
int n; /创建的矩阵为n*n/
int count = 1; /蛇形矩阵由1开始计数/
int round; /矩阵转的圈数/

int t,i;
scanf("%d",&t);
for(i=0;i<t;i++)

{
    scanf("%d",&n);

    count=1;
     int (*a)[n] = calloc(n * n,sizeof(int));

             //如果n = 1,则直接输出
    if(n == 1)
    {
        a[0][0] = count;
    }
    else
    {
        for(round = 0;round <= n / 2;round++)
        {
            x = round;
            for(y = round;y < n - round;y++)
            {
                a[x][y] = count;
                count++;
            }

            y = n - round - 1;
            for(x = round + 1;x < n - round - 1;x++)
            {
                a[x][y] = count;
                count++;
            }

            x = n - round - 1;
            for(y = n - round - 1;y >= round;y--)
            {
                a[x][y] = count;
                count++;
            }

            y = round;
            for(x = n - round - 2;x > round;x--)
            {
                a[x][y] = count;
                count++;
            }
        }

        if(n % 2 == 1)
        {
            a[n / 2][n / 2] = n*n;       //n为奇数时,“最中间的
        }
    }
    printf("case #%d:\n",i);

    for(x = 0;x< n;x++)
    {
        for(y = 0;y < n;y++)
        {
            printf("%d",a[x][y]);
            if(y<n-1) printf(" ");
        }
        printf("\n");
    }

}
return 0;

}

Li Dao

题解
写错了,是EOJ 2979

Li Dao

题解

include

using namespace std;
const int dx[4]={0,1,0,-1};
const int dy[4]={1,0,-1,0};
int T,n,tot,d;
int tu[20][20];

int isout(int xx)
{
if(xx<1 || xx>n) return 1;
else return 0;
}
int can_write(int xx,int yy)
{
if(!isout(xx) && !isout(yy) && tu[xx][yy]==0) return 1;
else return 0;
}
void write(int& xx,int& yy)
{
tu[xx][yy]=++tot;
if(tot==n*n) return;

int fx=xx+dx[d],fy=yy+dy[d];
while(!can_write(fx,fy))
{
d=(d+1)%4;
fx=xx+dx[d],fy=yy+dy[d];
}
xx=fx;yy=fy;return;
}
int main()
{
cin>>T;
for(int step=0;step>n;
int xx=1,yy=1;
d=0;
tot=0;
memset(tu,0,sizeof(tu));

for(int i=1;i<=n*n;i++) write(xx,yy);
printf("case #%d:\n",step);
for(int i=1;i<=n;i++)
{
  cout<<tu[i][1];
  for(int j=2;j<=n;j++) cout<<' '<<tu[i][j];
  cout<<endl;
}

}
return 0;
}

直接模拟,姊妹题:EOJ 2983

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