链接：https://ac.nowcoder.com/acm/contest/560/D
来源：牛客网

题目描述

Give you a rectangular gird which is h cells high and w cells wide.
Each grid is black initially. You can turn some grids into white.
A grid A(x,y) is connected with grid B if the coordinate of B is (x+1, y),(x-1, y),(x, y+1) or (x, y-1).
And your task is to propose a plan of the gird which has exactly n connected components of black part.
If there is no valid plan containing n connected components of black part, output -1.

输入描述:

Three integers h, w, n(1≤h,w≤200,1≤n≤109)(1≤h,w≤200,1≤n≤109) as described above.

输出描述:

Print h rows and w columns, '#' represents a black grid and '*' represents a white grid, indicating your solution.
示例1

输入

复制 
1 10 5

输出

复制 
#*#*#*#*#*

题意：
给你一个高h，宽w，一个数量k。让你构建一个h*w的数组，使之只含有两种元素，黑和白，即#和*
要求黑色块的联通集个数刚好是K。

思路：显然每一行中交叉填黑白，换行后每一行交叉填白黑，这样可以让这个h*w的数组中出现最大数量的黑色的联通块。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll h,w,n;
int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    gbtb;
    cin>>h>>w>>n;
    ll cnt=0ll;
    int flag=1;
    repd(i,1,h)
    {
        if(flag)
        {
            cnt+=(w+1)/2;
        }else
        {
            cnt+=(w-1)/2;
        }
        flag=!flag;
    }
    if(cnt>=n)
    {
        flag=1;
        repd(i,1,h)
        {
            if(flag)
            {
                repd(j,1,w)
                {
                    if(j&1)
                    {
                        if(n>0)
                        {
                            cout<<"#";
                            n--;
                        }else
                        {
                            cout<<"*";
                        }

                    }else
                    {
                        cout<<"*";
                    }
                }
                cout<<endl;
            }else
            {
                repd(j,1,w)
                {
                    if(j&1)
                    {
                        cout<<"*";
                    }else
                    {
                        if(n>0)
                        {
                            cout<<"#";
                            n--;
                        }else
                        {
                            cout<<"*";
                        }

                    }
                }
                cout<<endl;
            }
            flag=!flag;
        }
    }else
    {
        cout<<-1<<endl;
    }

    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}