【CodeforcesRound#275(Div.2)a_html/css_WEB-ITnose】教程文章相关的互联网学习教程文章

CodeforcesRound#250Div.2(C.TheChildandToy)_html/css_WEB-ITnose

题目如下: C. The Child and Toy time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output On Childrens Day, the child got a toy from Delayyy as a present. However, the child is so naughty that he cant wait to destroy the toy. The toy consists of n parts ...

CodeforcesRound#252(Div.2)-C,D_html/css_WEB-ITnose

C题就是一个简单的模拟,首先给每个人两个。然后把剩下的都给一个人就好了。 给的时候蛇形给。 #include#include#include#include#include#includeusing namespace std;#define LL __int64#define maxn 330000int main(){ int n,m,k; while(~scanf("%d%d%d",&n,&m,&k)) { int leap=1; int stx=1; int sty=1; int ms=n*m-(k*2)+2; printf("%d",ms); while(ms--) { ...

CodeforcesRound#251(Div.2)-C,D_html/css_WEB-ITnose

C题: 思路很简单。 由题意可知,奇数的集合有k-p个,偶数的几何有p个。 那么我们先选择k-p-1个奇数,每一个奇数是一个集合。 然后我们再选择p个偶数,每一个偶数是一个集合,如果偶数的个数不足,那么就用两个奇数去补。 然后我们再把剩下的所有的数放在一个集合里。 By rowanhao, contest: Codeforces Round #251 (Div. 2), problem: (C) Devu and Partitioning of the Array, Accepted, # #include#include#include#inc...

CodeforcesRound#253(Div.1)-A,B_html/css_WEB-ITnose

A题: 由题意可知,最多翻10次就可以(其实8次就够了),那么我们就用状态压缩表示状态。 对于某种状态,如果某一位为0,那么代表这一位不翻,否则代表这一位翻。 对于某一种翻的状态: 如果牌中有G3,那么就把G和3进行连边。其他的连边类似,不要重边。 对于任意一条边的两个端点,分三种情况讨论: 1,两个端点都翻了,那么很明显,这张牌被表示出来了。 2,两个端点中只有一个端点被翻,那么这个对应的num加1. 3,两个端...

CodeforcesRound#256(Div.2)C.PaintingFence(分治贪心)_html/css_WEB-ITnose

题目链接:http://codeforces.com/problemset/problem/448/C ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943/main -----------------------------------------------------------------------------------------------------------------------------...

CodeforcesRound#FF(Div.2)D.DZYLovesModification贪心+优先队列_html/css_WEB-ITnose

链接:http://codeforces.com/problemset/problem/447/D 题意:一个n*m的矩阵,可以进行k次操作,每次操作室对某一行或某一列的的数都减p,获得的得分是这一行或列原来的数字之和。求N次操作之后得到的最高得分是多少。 思路:首先分别统计每行和每列的数字和。 进行的k次操作中,有i次操作是对行进行操作,剩余k-i次操作是对列进行操作。首先在操作中忽略每次操作中行对列的影响,然后计算列的时候,最后可以计算出,总共的影响...

CodeforcesRound#256(Div.2)题解_html/css_WEB-ITnose

Problem A: A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present ? a new glass cupboard with n shelves and he decided to put all his presen...

CodeforcesRound#248(Div.1)??Nanami'sDigitalBoard_html/css_WEB-ITnose

题目连接 题意: 给n*m的0/1矩阵,q次操作,每次有两种:1)将x,y位置值翻转 2)计算以(x,y)为边界的矩形的面积最大值 (1?≤?n,?m,?q?≤?1000) 分析: 考虑以(x,y)为下边界的情况,h=(x,y)上边最多的连续1的个数。那么递减的枚举,对于当前hx,只需要看两侧能到达的最远距离,使得h(x,ty)不大于h即可。之后的枚举得到的两侧距离大于等于之前的,所以继续之前的两侧距离继续枚举即可。 const int maxn = 1100...

CodeforcesRound#256(Div.2)D.MultiplicationTable_html/css_WEB-ITnose

题目链接:http://codeforces.com/contest/448/problem/D 思路:用二分法 code: #include#include#includeusing namespace std;__int64 n,m,k;__int64 f(__int64 x){ __int64 res=0; for(__int64 i=1;i<=n;i++) { __int64 minn=min(m,x/i); //计算第i行有多少个数比x小,并且最多也只要m个数比x小 res+=minn; //计算出比x小的数的共有多少个 } return res<k;}int main(){ w...

CodeforcesRound#234(Div.2):B.InnaandNewMatrixofCandies_html/css_WEB-ITnose

B. Inna and New Matrix of Candies time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Inna likes sweets and a game called the "Candy Matrix". Today, she came up with the new game "Candy Matrix 2: Reload". The field for the new game is a rectangle table...

CodeforcesRound#256(Div.2)B.SuffixStructures(模拟)_html/css_WEB-ITnose

题目链接:http://codeforces.com/contest/448/problem/B ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943/main -----------------------------------------------------------------------------------------------------------------------------...

CodeforcesRound#256(Div.2)D二分答案_html/css_WEB-ITnose

D. Multiplication Table time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion isnt just charming, he also is very smart. While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champio...

CodeforcesRound#258(Div.2)B.JzzhuandSequences(矩阵快速幂)_html/css_WEB-ITnose

题目链接:http://codeforces.com/problemset/problem/450/B ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943/main -----------------------------------------------------------------------------------------------------------------------------...

CodeforcesRound#256(Div.2)A.Rewards_html/css_WEB-ITnose

A. Rewards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bizon the Champion is called the Champion for a reason. Bizon the Champion has recently got a present ? a new glass cupboard with n shelves and he decided to put all his presents there. All the pr...

CodeforcesRound#257(Div.2)A.JzzhuandChildren(简单题)_html/css_WEB-ITnose

题目链接:http://codeforces.com/problemset/problem/450/A ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943/main --------------------------------------------------------------------------------------------------------------------------...