CodeForces 1551D2 : Domino (hard version) 模拟

传送门

题意

给你一个 n ∗ m n * m n∗m矩阵，需要放若干个多米诺骨牌，要求其中水平的有 k k k个

分析

锻炼一下模拟能力吧

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 1e3 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
	char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
	while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int n,m,k;
int a[N][N];

int main() {
	int T;
	read(T);
	while(T--){
		memset(a,-1,sizeof a);
		read(n),read(m),read(k);
		bool flag = false;
		if(n * m < k * 2) puts("NO"),flag = true;
		else if(m % 2){
			if((m - 1) * n < k * 2) puts("NO"),flag = true;
			else if(k % 2 == 0) puts("YES");
			else puts("NO"),flag = true;
		}
		else if(n % 2){
			if((k - m / 2) >= 0 && (k - m / 2) % 2 == 0) puts("YES");
			else puts("NO"),flag = true;
		}
		else{
			if(k & 1) puts("NO"),flag = true;
			else puts("YES");
		}
		if(flag) continue;
		if(m % 2){
			for(int i = 1;i < m && k;i += 2){
				for(int j = 1;j <= n && k;j++,k--){
					int x = a[j - 1][i] + 1;
					if(x == a[j][i - 1]) x++;
					x %= 26;
					a[j][i] = a[j][i + 1]=  x;
				}
			}
			for(int i = 1;i <= n;i += 2)
				for(int j = 1;j <= m;j++){
					if(a[i][j] != -1) continue;
					int x = a[i - 1][j] + 1;
					if(x == a[i][j - 1]) x++;
					x %= 26;
					if(x == a[i + 1][j - 1])x++;
					x %= 26;
					a[i][j] = a[i + 1][j] = x;
				}
		}
		else if(n % 2){
			for(int i = 1;i <= m;i += 2,k--){
				int x = a[1][i - 1] + 1;
				x++;
				x %= 26;
				a[1][i] = a[1][i + 1] = x;
			}
			for(int i = 1;i <= m && k;i += 2){
				for(int j = 1;j <= n && k;j++){
					if(a[j][i] != -1) continue;
					k--;
					int x = a[j - 1][i] + 1;
					if(x == a[j][i - 1]) x++;
					x %= 26;
					a[j][i] = a[j][i + 1]=  x;
				}
			}
			for(int i = 2;i <= n;i += 2)
				for(int j = 1;j <= m;j++){
					if(a[i][j] != -1) continue;
					int x = a[i - 1][j] + 1;
					if(x == a[i][j - 1]) x++;
					x %= 26;
					if(x == a[i + 1][j - 1])x++;
					x %= 26;
					a[i][j] = a[i + 1][j] = x;
				}
		}
		else{
			for(int i = 1;i <= m && k;i += 2){
				for(int j = 1;j <= n && k;j++,k--){
					int x = a[j - 1][i] + 1;
					if(x == a[j][i - 1]) x++;
					x %= 26;
					a[j][i] = a[j][i + 1]=  x;
				}
			}
			for(int i = 1;i <= n;i += 2)
				for(int j = 1;j <= m;j++){
					if(a[i][j] != -1) continue;
					int x = a[i - 1][j] + 1;
					if(x == a[i][j - 1]) x++;
					x %= 26;
					if(x == a[i + 1][j - 1])x++;
					x %= 26;
					a[i][j] = a[i + 1][j] = x;
				}
		}
		for(int i = 1;i <= n;i++){
			for(int j = 1;j <= m;j++) printf("%c",a[i][j] + 'a');
			puts("");
		}
	}
	return 0;
}

/**
*　　┏┓　　　┏┓+ +
*　┏┛┻━━━┛┻┓ + +
*　┃　　　　　　　┃
*　┃　　　━　　　┃ ++ + + +
*  ████━████+
*  ◥██◤　◥██◤ +
*　┃　　　┻　　　┃
*　┃　　　　　　　┃ + +
*　┗━┓　　　┏━┛
*　　　┃　　　┃ + + + +Code is far away from 　
*　　　┃　　　┃ + bug with the animal protecting
*　　　┃　 　 ┗━━━┓ 神兽保佑,代码无bug　
*　　　┃ 　　　　　　 ┣┓
*　　  ┃ 　　　　　 　┏┛
*　    ┗┓┓┏━┳┓┏┛ + + + +
*　　　　┃┫┫　┃┫┫
*　　　　┗┻┛　┗┻┛+ + + +
*/

标签：26,puts,int,Domino,hard,1551D2,else,++,flag
来源： https://blog.csdn.net/tlyzxc/article/details/119063735