$Sol$
一年前做过差不多的南蛮图腾,当时做出来还是很有成就感的$OvO$
$N<=7$,就是模拟模拟,预处理一下,$over$
$Code$
#include<bits/stdc++.h> #define il inline #define Rg register #define go(i,a,b) for(Rg int i=a;i<=b;++i) #define yes(i,a,b) for(Rg int i=a;i>=b;--i) #define mem(a,b) memset(a,b,sizeof(a)) #define ll long long #define db double using namespace std; il int read() { Rg int x=0,y=1;char c=getchar(); while(c<‘0‘||c>‘9‘){if(c==‘-‘)y=-1;c=getchar();} while(c>=‘0‘&&c<=‘9‘){x=(x<<1)+(x<<3)+c-‘0‘;c=getchar();} return x*y; } char as[8][730][730]; il int ksm(int x,int y){Rg int ret=1;while(y){if(y&1)ret*=x;x*=x;y>>=1;}return ret;} il void init() { as[1][1][1]=‘X‘; go(n,2,7) { Rg int qwq=ksm(3,n-2); go(i,1,qwq) { go(j,1,qwq)as[n][i][j]=as[n][i][j+qwq*2]=as[n-1][i][j]; go(j,qwq+1,2*qwq)as[n][i][j]=‘ ‘; } go(i,qwq+1,qwq*2) { go(j,1,qwq)as[n][i][j]=as[n][i][j+qwq*2]=‘ ‘; go(j,qwq+1,2*qwq)as[n][i][j]=as[n-1][i-qwq][j-qwq]; } go(i,qwq*2+1,qwq*3) { go(j,1,qwq)as[n][i][j]=as[n][i][j+qwq*2]=as[n-1][i-qwq*2][j]; go(j,qwq+1,2*qwq)as[n][i][j]=‘ ‘; } } } int main() { init(); while(1) { Rg int n=read();if(n==-1)break; Rg int qwq=ksm(3,n-1); go(i,1,qwq){go(j,1,qwq)printf("%c",as[n][i][j]);printf("\n");} printf("-\n"); } return 0; }
原文地址:https://www.cnblogs.com/forward777/p/11279274.html
时间: 2024-10-07 23:05:46