Zoj--3839（模拟）

2014-11-30 20:10:06

思路：zoj11月月赛签到题，一开始脑抽写了200+行WA，精简到100行左右A掉了。其实就是构造，算好第i幅图再去算i+1幅图，计算好几个关键点很重要。

  1 /*************************************************************************
  2     > File Name: J.cpp
  3     > Author: Nature
  4     > Mail: 564374850@qq.com
  5     > Created Time: Sun 30 Nov 2014 01:41:11 PM CST
  6 ************************************************************************/
  7 
  8 #include <cstdio>
  9 #include <cstring>
 10 #include <cstdlib>
 11 #include <cmath>
 12 #include <vector>
 13 #include <map>
 14 #include <set>
 15 #include <stack>
 16 #include <queue>
 17 #include <iostream>
 18 #include <algorithm>
 19 using namespace std;
 20 #define lp (p << 1)
 21 #define rp (p << 1|1)
 22 #define getmid(l,r) (l + (r - l) / 2)
 23 #define MP(a,b) make_pair(a,b)
 24 typedef long long ll;
 25 const int INF = 1 << 30;
 26 
 27 char g3[20][20] = {"********","***  ***","***  ***","***  ***","* **** *",
 28     "* *  * *","* *  * *","********"};
 29 char g[7][1030][1030];
 30 int n;
 31 int m,A,B,C,a,b,c,d,e,f;
 32 
 33 void Mod(int k){
 34     int mm = 1;
 35     while(k--) mm *= 2;
 36     A = a = mm / 8;
 37     B = A + mm / 4;
 38     C = mm / 2;
 39     b = B - 1;
 40     c = mm - B;
 41     d = mm - 1 - a;
 42     e = mm / 4;
 43     f = mm - 1 - e;
 44 }
 45 
 46 int main(){
 47     for(int tt = 4; tt <= 10; ++tt){
 48         Mod(tt);
 49         m = 1 << tt;
 50         int t = tt - 4;
 51         for(int i = 0; i < m; ++i) g[t][0][i] = g[t][m - 1][i] = '*';
 52         for(int i = 1; i < m - 1; ++i){
 53             g[t][i][0] = g[t][i][m - 1] = '*';
 54             for(int j = 1; j < m - 1; ++j) g[t][i][j] = ' ';
 55         }
 56         for(int i = a; i <= b; ++i) g[t][A][i] = g[t][B][i] = '*';
 57         for(int i = c; i <= d; ++i) g[t][A][i] = g[t][B][i] = '*';
 58         for(int i = A; i <= B; ++i) g[t][i][a] = g[t][i][b] = g[t][i][c] = g[t][i][d] = '*';
 59         if(t == 0){
 60             for(int i = C; i < m; ++i){
 61                 for(int j = e; j <= f; ++j){
 62                     g[t][i][j] = g3[m / 2 - 1 - (i - C)][j - e];
 63                 }
 64             }
 65         }
 66         else{
 67             for(int i = C; i < m; ++i){
 68                 for(int j = e; j <= f; ++j){
 69                     g[t][i][j] = g[t - 1][m / 2 - 1 - (i - C)][j - e];
 70                 }
 71             }
 72         }
 73     }
 74     while(scanf("%d",&n) != EOF){
 75         if(n < 8) break;
 76         if(n == 8)
 77             for(int i = 0; i < n; ++i){
 78                 for(int j = 0; j < n; ++j){
 79                     printf("%c",g3[i][j]);
 80                 }
 81                 puts("");
 82             }
 83         else{
 84             int cnt = 0,tmp = n;
 85             while(tmp != 1){
 86                 cnt++;
 87                 tmp /= 2;
 88             }
 89             cnt -= 4;
 90             for(int i = 0; i < n; ++i){
 91                 for(int j = 0; j < n; ++j){
 92                     printf("%c",g[cnt][i][j]);
 93                 }
 94                 puts("");
 95             }
 96         }
 97         puts("");
 98     }
 99     return 0;
100 }

 

posted @ 2014-11-30 20:12  Naturain  阅读(162)  评论(0)  编辑  收藏  举报