zoj 3212 K-Nice(数学题)
来源:互联网 发布:努比亚手机怎么样 知乎 编辑:程序博客网 时间:2024/05/24 06:27
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3212
转载请注明出处:http://blog.csdn.net/u012860063
This is a super simple problem. The description is simple, the solution is simple. If you believe so, just read it on. Or if you don't, just pretend that you can't see this one.
We say an element is inside a matrix if it has four neighboring elements in the matrix (Those at the corner have two and on the edge have three). An element inside a matrix is called "nice" when its value equals the sum of its four neighbors. A matrix is called "k-nice" if and only if k of the elements inside the matrix are "nice".
Now given the size of the matrix and the value of k, you are to output any one of the "k-nice" matrix of the given size. It is guaranteed that there is always a solution to every test case.
Input
The first line of the input contains an integer T (1 <= T <= 8500) followed by T test cases. Each case contains three integers n, m, k (2 <= n, m <= 15, 0 <= k <= (n - 2) * (m - 2)) indicating the matrix size n * m and it the "nice"-degree k.
Output
For each test case, output a matrix with n lines each containing m elements separated by a space (no extra space at the end of the line). The absolute value of the elements in the matrix should not be greater than 10000.
Sample Input
24 5 35 5 3
Sample Output
2 1 3 1 14 8 2 6 11 1 9 2 92 2 4 4 30 1 2 3 00 4 5 6 00 0 0 0 00 0 0 0 00 0 0 0 0
Author: ZHUANG, Junyuan
Source: The 6th Zhejiang Provincial Collegiate Programming Contest
题意:
任意输出一个矩阵要求其中有且仅有K个满足题意的点!
满足要求的点即要此点的上下左右的和等于此点!
代码如下:
#include<cstdio>int map[22][22];void bai(int x,int y){map[x][y+1]=0;map[x][y-1]=0;map[x+1][y]=0;map[x-1][y]=0;}int main(){int t,x,y,k,i,j;scanf("%d",&t);while(t--){scanf("%d%d%d",&x,&y,&k);for(i=0;i<=x+1;i++){for(j=0;j<=y+1;j++)map[i][j]=1;}int count=0,flag=0;if(k!=0){for(i=2;i<=x-1;i++){for(j=2;j<=y-1;j++){map[i][j]=0;bai(i,j);count++;if(count==k){flag=1;break;}}if(flag==1)break;}}for(i=1;i<=x;i++){printf("%d",map[i][1]);for(j=2;j<=y;j++){printf(" %d",map[i][j]);}printf("\n");}}return 0;}
- zoj 3212 K-Nice(数学题)
- ZOJ - 3212 K-Nice
- zoj 3212 K-Nice
- ZOJ 3212 K-Nice
- ZOJ 3212 K-Nice
- ZOJ 3212 K-Nice【】
- ZOJ 3212 K-Nice
- ZOJ 3212 K-Nice(思维)
- ZOJ 3212K-Nice(构造)
- zoj 3212 K-Nice 解题报告
- 【城会玩系列】ZOJ 3212 K-Nice【思维】
- ZOJ-3212-K-Nice【6th浙江省赛】【构造】
- ZOJ3206 Disaster Area Reconstruction ZOJ 3211 Dream City ZOJ 3212 K-Nice
- ZOJ 3212(K)
- K-Nice
- K-Nice
- ZOJ 1318 (第k个序列,简单数学题)
- zoj 2975 Kinds of Fuwas(数学题)
- 网络爬虫初级
- 李成名:科学就是较真 数字城市/智慧城市就是跑马圈地
- ASP.NET之数据库设计与SQL语句(旗舰版1)
- Android 布局之layout
- 这尼玛都太逗了吧
- zoj 3212 K-Nice(数学题)
- if标签的分析
- C++析构函数为什么要为虚函数
- 站帮网微管家 为你打通微信和Discuz论坛社区
- 月不明,天尤暗,春水潺潺,雨丝斑斑
- strcpy实现
- ios basic:1.1 design a user interface(学习在stoaryboard上build and manage views )
- PL/SQL Developer如何连接64位的Oracle图解
- 数据结构 queque