Codeforces Round #342 (Div. 2) C
2 seconds
256 megabytes
standard input
standard output
People do many crazy things to stand out in a crowd. Some of them dance, some learn by heart rules of Russian language, some try to become an outstanding competitive programmers, while others collect funny math objects.
Alis is among these collectors. Right now she wants to get one of k-special tables. In case you forget, the table n × n is called k-special if the following three conditions are satisfied:
- every integer from 1 to n2 appears in the table exactly once;
- in each row numbers are situated in increasing order;
- the sum of numbers in the k-th column is maximum possible.
Your goal is to help Alice and find at least one k-special table of size n × n. Both rows and columns are numbered from 1 to n, with rows numbered from top to bottom and columns numbered from left to right.
The first line of the input contains two integers n and k (1 ≤ n ≤ 500, 1 ≤ k ≤ n) — the size of the table Alice is looking for and the column that should have maximum possible sum.
First print the sum of the integers in the k-th column of the required table.
Next n lines should contain the description of the table itself: first line should contains n elements of the first row, second line should contain n elements of the second row and so on.
If there are multiple suitable table, you are allowed to print any.
4 1
28
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
5 3
85
5 6 17 18 19
9 10 23 24 25
7 8 20 21 22
3 4 14 15 16
1 2 11 12 13 题意:1~n*n 排列在n*n的矩阵 要求每个数字出现一次
每行递增 第k列的和尽可能的大
题解:模拟 以第k列为分界线 左边为 1~n*(k-1)的排列
右边为n*(k-1)+1~n*n 的排列
简单模拟
#include<iostream>
#include<cstdio>
#include<cstring>
#define LL __int64
using namespace std;
int n,k;
int main()
{
scanf("%d%d",&n,&k);
int exm;
int ans=0;
exm=n*(k-1);
int gg=1,ggg=exm+1;
for(int i=exm+1;i<=n*n;i=i+n-k+1)
ans+=i;
printf("%d\n",ans);
for(int i=1;i<=n;i++)
{
for(int j=1;j<k;j++)
{
if(j==1)
printf("%d",gg);
else
printf(" %d",gg);
gg++;
}
for(int j=k;j<=n;j++)
{
if(j==1)
printf("%d",ggg);
else
printf(" %d",ggg);
ggg++;
}
printf("\n");
}
return 0;
}
Codeforces Round #342 (Div. 2) C的更多相关文章
- Codeforces Round #342 (Div. 2) D. Finals in arithmetic 贪心
D. Finals in arithmetic 题目连接: http://www.codeforces.com/contest/625/problem/D Description Vitya is s ...
- Codeforces Round #342 (Div. 2) C. K-special Tables 构造
C. K-special Tables 题目连接: http://www.codeforces.com/contest/625/problem/C Description People do many ...
- Codeforces Round #342 (Div. 2) B. War of the Corporations 贪心
B. War of the Corporations 题目连接: http://www.codeforces.com/contest/625/problem/B Description A long ...
- Codeforces Round #342 (Div. 2) A - Guest From the Past 数学
A. Guest From the Past 题目连接: http://www.codeforces.com/contest/625/problem/A Description Kolya Geras ...
- Codeforces Round #342 (Div. 2) E. Frog Fights set 模拟
E. Frog Fights 题目连接: http://www.codeforces.com/contest/625/problem/E Description stap Bender recentl ...
- Codeforces Round #342 (Div. 2) D. Finals in arithmetic(想法题/构造题)
传送门 Description Vitya is studying in the third grade. During the last math lesson all the pupils wro ...
- Codeforces Round #342 (Div. 2) C. K-special Tables(想法题)
传送门 Description People do many crazy things to stand out in a crowd. Some of them dance, some learn ...
- Codeforces Round #342 (Div. 2) B. War of the Corporations(贪心)
传送门 Description A long time ago, in a galaxy far far away two giant IT-corporations Pineapple and Go ...
- Codeforces Round #342 (Div. 2) A. Guest From the Past(贪心)
传送门 Description Kolya Gerasimov loves kefir very much. He lives in year 1984 and knows all the detai ...
- Codeforces Round #342 (Div. 2)-B. War of the Corporations
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...
随机推荐
- PAT - L2-001. 紧急救援( Dijstra )
- PAT - L2-001. 紧急救援 题目链接 作为一个城市的应急救援队伍的负责人,你有一张特殊的全国地图.在地图上显示有多个分散的城市和一些连接城市的快速道路.每个城市的救援队数量和每一条连接两 ...
- system_Class类说明文档
system_Class类是FastCMS系统必须的,全局对象system是system_Class的实例,其主要包含二类操作: 1.token 操作: token可以存储当前访客的私有信息,取代se ...
- vuex-Actions的用法
Action 类似于 mutation,不同在于: Action 提交的是 mutation,而不是直接变更状态. Action 是异步的,mutation是同步的. 沿用vuex学习---简介的案例 ...
- 打印队列 (Printer Queue,ACM/ICPC NWERC 2006,UVA12100)
题目描述: 题目思路: 使用一个队列记录数字,一个优先队列记录优先级,如果相等即可打印: #include <iostream> #include <queue> using ...
- avalonJS入门
前端神器avalonJS入门(一) posted @ 2014-10-31 17:44 vajoy 阅读(8759) 评论(42) 编辑 收藏 avalonJS是司徒正美开发和维护的前端mvvm框 ...
- 阿里校招内推C++岗位编程题第一题 空格最少的字符串
给定一个字符串S和有效单词的字典D,请确定可以插入到S中的最小空格数,使得最终的字符串完全由D中的有效单词组成.并输出解. 如果没有解则应该输出n/a 例如: 输入: S = “ilikealibab ...
- POJ 2631 Roads in the North(求树的直径,两次遍历 or 树DP)
题目链接:http://poj.org/problem?id=2631 Description Building and maintaining roads among communities in ...
- JS原型与面向对象总结
ECMAScript有两种开发模式:1.函数式(过程化),2.面向对象(OOP).面向对象的语言有一个标志,那就是类的概念,而通过类可以创建任意多个具有相同属性和方法的对象.但 是,ECMAScrip ...
- PHPCMS调取当前栏目的描述、文章位置导航、当前栏目链接、当前栏目名称
当我们填写了栏目描述,怎么调用出来. 使用 {$CATEGORYS[$catid][description]} 就可以把栏目的描述调用出来 下面三个也比较常用{catpos($catid)} 显示文章 ...
- 2d命令行小游戏源码
using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.T ...