PAT (Advanced Level) 1105. Spiral Matrix (25)
简单模拟。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<queue>
#include<string>
#include<algorithm>
using namespace std; const int maxn=+;
int N,a[maxn];
int ans[maxn];
int m,n;
int dir[][]; void f()
{
dir[][]=;dir[][]=;
dir[][]=;dir[][]=;
dir[][]=;dir[][]=-;
dir[][]=-;dir[][]=;
} bool cmp(const int&a,const int&b)
{
return a>b;
} int main()
{
scanf("%d",&N);
f();
for(int i=;i<=N;i++) scanf("%d",&a[i]);
sort(a+,a++N,cmp); for(int i=sqrt(1.0*N);i>=;i--)
if(N%i==){n=i,m=N/i;break;} memset(ans,-,sizeof ans);
for(int i=;i<=m;i++)
for(int j=;j<=n;j++)
ans[(i-)*n+j]=; int nowDir=;
int nowX=,nowY=;
for(int i=;i<=N;i++)
{
ans[(nowX-)*n+nowY]=a[i];
int NewX=nowX+dir[nowDir][];
int NewY=nowY+dir[nowDir][];
if(ans[(NewX-)*n+NewY]!=||NewX<=||NewX>=m+||NewY<=||NewY>=n+) nowDir=(nowDir+)%;
nowX=nowX+dir[nowDir][];
nowY=nowY+dir[nowDir][];
}
int num=;
// for(int i=1;i<=m*n;i++) printf("%d ",ans[i]); for(int i=;i<=m;i++)
for(int j=;j<=n;j++)
{
printf("%d",ans[num++]);
if(j<n) printf(" ");
else printf("\n");
}
return ;
}
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