题解

34 条题解

  • -1
    @ 2015-11-14 15:37:08

    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #define mid k/2+1
    int ans[40][40],p_l[1600],p_r[1600],have[40][40];
    void solve(int k){
    int i;
    for (i=2; i<=k*k; i++) {
    if (p_l[i-1]==1&&p_r[i-1]!=k) {
    p_l[i]=k;
    p_r[i]=p_r[i-1]+1;
    have[p_l[i]][p_r[i]]=1;
    }else if(p_r[i-1]==k&&p_l[i-1]!=1){
    p_r[i]=1;
    p_l[i]=p_l[i-1]-1;
    have[p_l[i]][p_r[i]]=1;
    }else if(p_l[i-1]==1&&p_r[i-1]==k){
    p_l[i]=p_l[i-1]+1;
    p_r[i]=p_r[i-1];
    have[p_l[i]][p_r[i]]=1;
    }else{
    if (have[p_l[i-1]-1][p_r[i-1]+1]==0) {
    p_l[i]=p_l[i-1]-1;
    p_r[i]=p_r[i-1]+1;
    have[p_l[i]][p_r[i]]=1;
    }else{
    p_l[i]=p_l[i-1]+1;
    p_r[i]=p_r[i-1];
    have[p_l[i]][p_r[i]]=1;
    }
    }
    }
    }
    int main(){
    int k,i,j;
    scanf("%d",&k);
    memset(ans,0,sizeof(ans));
    memset(have,0,sizeof(have));
    if(k%2!=0){
    p_l[1]=1;
    p_r[1]=mid;
    have[p_l[1]][p_r[1]]=1;
    solve(k);
    for (i=1; i<=k*k; i++) {
    ans[p_l[i]][p_r[i]]=i;
    }
    for (i=1; i<=k; i++) {
    for (j=1; j<=k; j++) {
    printf("%d ",ans[i][j]);
    }
    printf("\n");
    }
    }
    return 0;
    }

  • -2
    @ 2017-02-10 18:07:24

    评测结果
    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 544 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 548 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 544 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 544 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 544 KiB, score = 10
    测试数据 #5: Accepted, time = 0 ms, mem = 544 KiB, score = 10
    测试数据 #6: Accepted, time = 0 ms, mem = 544 KiB, score = 10
    测试数据 #7: Accepted, time = 15 ms, mem = 544 KiB, score = 10
    测试数据 #8: Accepted, time = 0 ms, mem = 544 KiB, score = 10
    测试数据 #9: Accepted, time = 15 ms, mem = 544 KiB, score = 10
    Accepted, time = 30 ms, mem = 548 KiB, score = 100
    代码

    include <cstdio>

    include <cstring>

    using namespace std;
    int main()
    {
    int n=0,h=0,l=0;
    int hf[41][41]={0};
    scanf("%d",&n);

    h=1;
    l=n/2+1;
    hf[h][l]=1;
    h=n;
    l++;
    for(long i=2;i<=n*n;i++)
    {
    if(h==1&&l!=n)
    {
    hf[h][l]=i;
    h=n;
    l++;
    continue;
    }
    if(l==n&&h!=1)
    {
    hf[h][l]=i;
    l=1;
    h--;
    continue;
    }
    if(h==1&&l==n)
    {
    hf[h][l]=i;
    h++;
    continue;
    }
    if(h!=1&&l!=n)
    {
    if(hf[h-1][l+1]!=0)
    {
    hf[h][l]=i;
    h++;
    continue;
    }
    if(hf[h-1][l+1]==0)
    {
    hf[h][l]=i;
    h--;
    l++;
    continue;
    }
    }
    }
    for(int i=1;i<=n;i++)
    {
    for(int j=1;j<=n;j++)
    {
    printf("%d ",hf[i][j]);
    }
    printf("\n");
    }
    return 0;
    }

  • -2
    @ 2016-11-07 14:21:16

    范围要大,几千才行
    不能用cout输出,要用printf

  • -2
    @ 2016-08-08 10:46:34

    var
    n,i,j:integer;
    a:array[0..39,0..39] of integer;

    procedure sub(k,x,y:integer); //搜索
    var u,v:integer;
    begin
    if k<=sqr(n) then begin
    if x=1 then begin
    if y=n then begin u:=x+1;v:=y;end
    else begin u:=n;v:=y+1;end;
    end
    else begin
    if y=n then begin v:=1;u:=x-1; end
    else begin
    if (a[x-1,y+1]=0) then begin u:=x-1;v:=y+1;end
    else begin u:=x+1;v:=y;end;
    end;
    end;

    a[u,v]:=k;
    sub(k+1,u,v);
    end;
    end;

    begin
    fillchar(a,sizeof(a),0);
    readln(n);

    a[1,(n+1) div 2]:=1;
    sub(2,1,(n+1) div 2);

    for i:=1 to n do
    begin
    for j:=1 to n do write(a[i,j],' ');
    writeln;
    end;

    end.

  • -2
    @ 2016-08-02 20:54:36

    注意空格啊= =醉了
    program magic;
    var k,x,y,i,j,n:integer;
    a:array [0..50,0..50] of integer;
    begin
    //assign
    //hang,lie
    read(n);
    fillchar(a,sizeof(a),0);
    a[1,n div 2 +1]:=1;
    x:=1; y:=n div 2 +1;
    for k:=2 to n*n do
    begin
    if (x=1) and (y<>n) then
    begin
    a[n,y+1]:=k;
    x:=n;
    y:=y+1;
    continue;
    end;
    if (x<>1) and (y=n) then
    begin
    a[x-1,1]:=k;
    x:=x-1;
    y:=1;
    continue;
    end;
    if (x=1) and (y=n) then
    begin
    a[x+1,y]:=k;
    x:=x+1;
    continue;
    end;
    if (x<>1) and (y<>n) then
    begin
    if (a[x-1,y+1]=0) then
    begin
    a[x-1,y+1]:=k;
    x:=x-1;
    y:=y+1;
    continue;
    end
    else begin a[x+1,y]:=k;
    x:=x+1;
    continue;
    end;

    end;
    end;
    for i:=1 to n do
    begin
    for j:=1 to n-1 do begin write(a[i,j],' '); end;
    writeln(a[i,n]);
    end;
    end.

  • -2
    @ 2016-07-20 21:33:47

    //c++模拟
    #include<iostream>

    #include<stdio.h>

    #include<string.h>

    using namespace std;

    int maze[50][50];
    int n;
    int main()
    {
    cin>>n;
    if(n==1)
    {
    cout<<1;
    return 0;
    }
    maze[1][(1+n)/2]=1;
    int hang=1;
    int lie=(1+n)/2;
    for(int k=2;k<=n*n;k++)
    {

    if((hang==1)&&(lie!=n))
    {
    hang=n;
    lie+=1;
    maze[hang][lie]=k;
    continue;
    }
    if((lie==n)&&(hang!=1))
    {
    lie=1;
    hang-=1;
    maze[hang][lie]=k;
    continue;
    }
    if(hang==1&&lie==n)
    {
    hang+=1;
    maze[hang][lie]=k;
    continue;
    }
    if(hang!=1)
    {
    if(lie!=n)
    {
    if(maze[hang-1][lie+1]==0)
    {
    hang-=1;
    lie+=1;
    maze[hang][lie]=k;
    continue;
    }
    else
    {
    hang+=1;
    maze[hang][lie]=k;
    continue;
    }
    }
    }
    ;
    }
    for(int i=1;i<=n;i++)
    {
    for(int j=1;j<=n;j++)
    {
    cout<<maze[i][j]<<" ";
    }
    cout<<endl;
    }
    return 0;

    }

  • -2
    @ 2016-07-12 12:24:32
    #include <bits/stdc++.h>
    #define MAX_N  50
    int arr[MAX_N + 10][MAX_N + 10];
    using namespace std;
    int n;
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(NULL);
        cin>>n;
        arr[1][(n + 1)>>1] = 1;
        int last_r = 1,last_c = ((n + 1)>>1),now_r = 0,now_c = 0;
        for (int i = 2,range=n*n; i <= range; i++) {
            if (last_r == 1 && last_c != n)
                now_r = n,now_c = last_c + 1;
            else if (last_r != 1 && last_c == n)
                now_r = last_r - 1,now_c = 1;
            else if (last_r == 1 && last_c == n)
                now_r = last_r + 1,now_c = last_c;
            else {
                if (!arr[last_r - 1][last_c + 1])
                    now_r = last_r - 1,now_c = last_c + 1;
                else
                    now_r = last_r + 1,now_c = last_c;
            }
            arr[now_r][now_c] = i,last_r = now_r,last_c = now_c;
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++) {
                if (j == 1)cout<<arr[i][j]<<" ";
                else cout<<arr[i][j]<<" ";
            }
            cout<<"\n";
        }
        return 0;
    }
    
  • -2
    @ 2015-11-14 18:52:20

    诶,退役了......浙江太NB,已炸

    这是道纯模拟题。模拟吧。没技巧,注意二维数组行和列。最好一开始就做个输出的子函数,每填一个数输出表格一次,可以加快排bug效率。

    pascal代码等我程序发回来再发...

  • -3
    @ 2017-05-18 13:42:43
    #include<stdio.h>
    int f[50][50],n,i,j,k=2;
    int main(){
        for(scanf("%d",&n),f[i=1][j=n+1>>1]=1;k<=n*n;f[i][j]=k++)
            if(i==1 && j<n) i=n,++j; else if(j==n&&i>1) j=1,i--;
            else if(i==1&&j==n) ++i; else !f[i-1][j+1]?i--,j++:i++;
        for(i=1;i<=n;++i,puts("")) for(j=0;j<n;printf("%d ",f[i][++j]));
    }
    
  • -3
    @ 2016-11-05 19:45:09

    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <cmath>
    #include <map>
    #include <vector>
    #include <stack>
    #include <queue>

    using namespace std;

    int f[1000][1000],n,x,y;

    int main(){
    scanf("%d",&n);
    f[x=1][y=(n+1)/2]=1;
    for(int k=2;k<=n*n;f[x][y]=k++){
    if(x==1&&y!=n)x=n,y=y+1;
    else if(y==n&&x!=1)y=1,x--;
    else if(x==1&&y==n)x++;
    else !f[x-1][y+1]&&(x--,y++)||x++;
    }
    for(int i=1;i<=n;i++,puts(""))
    for(int j=1;j<=n;printf("%d ",f[i][j++]));
    }

  • -3
    @ 2016-10-04 20:47:05

    可以用数学规律,第一行的正中间一定为1,然后向左移一格,向上移一格为下一个数,若到边界就回到另一边开始,若该格已有数则下移一格。

  • -3
    @ 2016-07-17 09:49:46

    这不就是硬解么...Pascal代码
    Pascal
    var n,x,y,k:integer;
    a:array[1..39,1..39] of integer;
    begin
    readln(n);fillchar(a,sizeof(a),0);
    y:=n div 2+1;x:=1;a[x,y]:=1;
    if n=1 then begin writeln('1');halt; end
    else begin
    for k:=2 to n*n do
    begin
    if (x=1) and (y<>n) then begin x:=n;inc(y); end
    else if (y=n) and (x<>1) then begin y:=1;dec(x); end
    else if (x=1) and (y=n) then inc(x)
    else if a[x-1,y+1]=0 then begin dec(X);inc(y); end
    else inc(x);
    a[x,y]:=k;
    end;
    end;
    for x:=1 to n do
    begin
    for y:=1 to n do write(a[x,y],' ');
    writeln;
    end;
    end.

  • -3
    @ 2016-06-19 10:41:11

    ~c++
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    const int maxn=40;
    int n,a[maxn][maxn],x,y;
    int main()
    {
    scanf("%d",&n);
    a[1][n/2+1]=1;

    x=1,y=(n/2)+1;
    for(int i=2;i<=n*n;i++)
    {
    if((x==1)&&(y!=n))x=n,y+=1;
    else
    if((x!=1)&&(y==n))y=1,x-=1;
    else
    if((x==1)&&(y==n))x=2,y=n;
    else
    if(a[x-1][y+1]==0)x-=1,y+=1;
    else
    x+=1,y=y;
    a[x][y]=i;

    }
    for(int i=1;i<=n;i++)
    {
    for(int j=1;j<=n;j++)printf("%d ",a[i][j]);
    printf("\n");

    }
    return 0;
    }

  • -4
    @ 2016-11-02 17:40:07
    #include <iostream>
    #include <cstring>
    using namespace std;
    struct _hash{
        int x,y;
    };
    int k;
    int map[40][40]={0};
    _hash myhash[40*40]={0};
        int n;
    void set(int k,int x,int y){
        myhash[k].x = x;
        myhash[k].y = y;
        map[x][y]=k;
    }
    int main(){
        cin>> n;
        memset(map,0,sizeof(map));
        memset(myhash,0,sizeof(myhash));
        set(1,1,n/2+1);
        for(k =2;k<=n*n;k++){
            if(myhash[k-1].x==1&&myhash[k-1].y!=n){
                set(k,n,myhash[k-1].y+1);
            }
            else if(myhash[k-1].y==n&&myhash[k-1].x!=1){
                set(k,myhash[k-1].x-1,1);
            }
            else if(myhash[k-1].x==1&&myhash[k-1].y==n){
                set(k,2,n);
            }
            else if(myhash[k-1].x!=1&&myhash[k-1].y!=n){
                
                if(map[myhash[k-1].x-1][myhash[k-1].y+1]==0){
                    set(k,myhash[k-1].x-1,myhash[k-1].y+1);
                }
                else{
                    set(k,myhash[k-1].x+1,myhash[k-1].y);
                }
            }
            else{}
        }
        for(int x=1;x<=n;x++){
            for(int y = 1;y<=n;y++){
                cout<<map[x][y]<<" ";
            }
            cout<<endl;
        }
        
        return 0;
    }
    

信息

ID
1978
难度
2
分类
(无)
标签
递交数
2438
已通过
1338
通过率
55%
被复制
8
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