HDOJ 2074 叠筐

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题意：实现如样例的图形，输入三个数据，分别为层数，中心图案，外框图案

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2074

思路：模拟水题，直接看代码吧。

注意点：注意两个图案的顺序。

以下为AC代码：

Run IDSubmit TimeJudge StatusPro.IDExe.TimeExe.MemoryCode Len.LanguageAuthor129761332015-02-22 15:53:15Accepted20740MS1304K2421 BG++luminous11

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define RDI(a) scanf ( "%d", &a )#define RDII(a, b) scanf ( "%d%d", &a, &b )#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c );#define RDS(s) scanf ( "%s", s );#define PIL(a) printf ( "%d\n", a );#define PIIL(a,b) printf ( "%d %d\n", a, b );#define PIIL(a,b,c) printf ( "%d %d %d\n", a, b, c );#define PSL(s) printf ( "%s\n", s );#define REP(i,m,n) for ( int i = m; i <= n; i ++ )#define DEP(i,m,n) for ( int i = m; i >= n; i -- )#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;const double pi = acos(-1);const double eps = 1e-10;const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };struct node{    int x, y, cnt;    node(){}    node( int _x, int _y ) : x(_x), y(_y) {}    node( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}};int n;string str_1, str_2;char str[100][1001];char a, b;void init(){    clr ( str, 0 );    a = str_1[0];    b = str_2[0];    if ( ( n - 1 ) % 4 == 0 )        swap ( a, b );}void solve(){    char tmp;    REP ( i, 0, ( n - 1 ) / 2 ){        if ( i & 1 )            tmp = a;        else            tmp = b;        REP ( j, i, n - 1 - i )            REP ( k, i, n - 1 - i )                str[j][k] = tmp;    }    if ( n != 1 )        str[0][0] = str[0][n-1] = str[n-1][0] = str[n-1][n-1] = ' ';    REP ( i, 0, n - 1 )        PSL ( str[i] );}int main(){    bool flag = 0;    while ( cin >> n >> str_1 >> str_2 ){        init();        if ( flag )puts("");        solve();        flag = 1;    }    return 0;}

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