Codeforces Round #280 (Div. 2) A B C 暴力 水 贪心
1 second
256 megabytes
standard input
standard output
Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube, the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes, and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.
Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.
The first line contains integer n (1 ≤ n ≤ 104) — the number of cubes given to Vanya.
Print the maximum possible height of the pyramid in the single line.
1
1
25
4
Illustration to the second sample:
题意:给你n个方块 按照如图的方式摆放 问最多能摆放多少层?
题解:暴力层数 判断需要多少方块
/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
#define ll __int64
using namespace std;
int n;
int a[];
int main()
{
scanf("%d",&n);
int sum=;
for(int i=;i<=;i++)
{
sum=sum+i*(i+)/;
if(sum==n)
{
cout<<i<<endl;
return ;
}
if(sum>n)
{
cout<<i-<<endl;
return ;
}
}
return ;
}
1 second
256 megabytes
standard input
standard output
Vanya walks late at night along a straight street of length l, lit by n lanterns. Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point l. Then the i-th lantern is at the point ai. The lantern lights all points of the street that are at the distance of at most d from it, where d is some positive number, common for all lanterns.
Vanya wonders: what is the minimum light radius d should the lanterns have to light the whole street?
The first line contains two integers n, l (1 ≤ n ≤ 1000, 1 ≤ l ≤ 109) — the number of lanterns and the length of the street respectively.
The next line contains n integers ai (0 ≤ ai ≤ l). Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.
Print the minimum light radius d, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10 - 9.
7 15
15 5 3 7 9 14 0
2.5000000000
2 5
2 5
2.0000000000
Consider the second sample. At d = 2 the first lantern will light the segment [0, 4] of the street, and the second lantern will light segment[3, 5]. Thus, the whole street will be lit.
题意:l长的路 给你n个路灯的坐标 问你最小的路灯照射半径使得路上都被照亮
题解:寻找最长的路灯间隔/2 并与第一个路灯,最后一个路灯照射的边界比较取最大值输出
/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
#define ll __int64
using namespace std;
int n,l;
double a[];
int main()
{
scanf("%d %d",&n,&l);
for(int i=;i<n;i++)
scanf("%lf",&a[i]);
sort(a,a+n);
double minx=;
for(int i=;i<n;i++)
{
if(a[i-]==a[i])
continue;
minx=max(minx,a[i]-a[i-]);
}
minx=max(minx/2.0,max(a[],l-a[n-]));
printf("%f\n",minx);
return ;
}
1 second
256 megabytes
standard input
standard output
Vanya wants to pass n exams and get the academic scholarship. He will get the scholarship if the average grade mark for all the exams is at least avg. The exam grade cannot exceed r. Vanya has passed the exams and got grade ai for the i-th exam. To increase the grade for the i-th exam by 1 point, Vanya must write bi essays. He can raise the exam grade multiple times.
What is the minimum number of essays that Vanya needs to write to get scholarship?
The first line contains three integers n, r, avg (1 ≤ n ≤ 105, 1 ≤ r ≤ 109, 1 ≤ avg ≤ min(r, 106)) — the number of exams, the maximum grade and the required grade point average, respectively.
Each of the following n lines contains space-separated integers ai and bi (1 ≤ ai ≤ r, 1 ≤ bi ≤ 106).
In the first line print the minimum number of essays.
5 5 4
5 2
4 7
3 1
3 2
2 5
4
2 5 4
5 2
5 2
0
In the first sample Vanya can write 2 essays for the 3rd exam to raise his grade by 2 points and 2 essays for the 4th exam to raise his grade by 1 point.
In the second sample, Vanya doesn't need to write any essays as his general point average already is above average.
题意:n门科目 每门满分r 要求所有科目的均分大于等于avg才能获得奖学金
对于每门科目都有两个值 a为得分 b为每增加一分需要看的文章的数量 输出想要获得奖学金 需要看的文章的数量的最小值
题解:贪心策略 对于每门科目按照b升序排
/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
#define ll __int64
using namespace std;
ll n,r,avg;
struct node
{
ll a,b;
}N[];
bool cmp(struct node aa,struct node bb)
{
return aa.b<bb.b;
}
int main()
{
scanf("%I64d %I64d %I64d",&n,&r,&avg);
ll sum=;
for(int i=;i<n;i++)
{
scanf("%I64d %I64d",&N[i].a,&N[i].b);
sum+=N[i].a;
}
if(avg*n<=sum)
{
printf("0\n");
return ;
}
sum=avg*n-sum;
sort(N,N+n,cmp);
ll ans=;
for(int i=;i<n;i++)
{
ll exm=N[i].a;
if(exm==r)
continue;
if(sum>=(r-exm))
{
ans=ans+(r-exm)*N[i].b;
sum=sum-(r-exm);
}
else
{
ans=ans+sum*N[i].b;
sum=;
}
if(sum==)
break;
}
printf("%I64d\n",ans);
return ;
}
Codeforces Round #280 (Div. 2) A B C 暴力 水 贪心的更多相关文章
- Codeforces Round #280 (Div. 2) A. Vanya and Cubes 水题
A. Vanya and Cubes time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #377 (Div. 2) A B C D 水/贪心/贪心/二分
A. Buy a Shovel time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题
Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec Memory Limit: 256 MBSubmit: xxx ...
- Codeforces Round #396 (Div. 2) A B C D 水 trick dp 并查集
A. Mahmoud and Longest Uncommon Subsequence time limit per test 2 seconds memory limit per test 256 ...
- Codeforces Round #280 (Div. 2) E. Vanya and Field 数学
E. Vanya and Field Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/492/pr ...
- Codeforces Round #280 (Div. 2) D. Vanya and Computer Game 二分
D. Vanya and Computer Game Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contes ...
- Codeforces Round #280 (Div. 2) C. Vanya and Exams 贪心
C. Vanya and Exams Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/492/pr ...
- CodeForces Round #280 (Div.2)
A. Vanya and Cubes 题意: 给你n个小方块,现在要搭一个金字塔,金字塔的第i层需要 个小方块,问这n个方块最多搭几层金字塔. 分析: 根据求和公式,有,按照规律直接加就行,直到超过n ...
- Codeforces Round #280 (Div. 2)E Vanya and Field(简单题)
转载请注明出处: http://www.cnblogs.com/fraud/ ——by fraud 本场题目都比较简单,故只写了E题. E. Vanya and Field Vany ...
随机推荐
- IBatis.Net 表连接查询(五)
IBatis.Net之多表查询: 一.定制实际对应类的方式 首先配置多表的测试数据库,IBatis.Net之Oracle表连接查询配置: 首先新建两张表如下: 为两张表建立外键: ALTER TABL ...
- 【C语言学习】-05 二维数组、字符串数组、多维数组
⼆二维数组.字符串数组.多维数组
- Android 圆形ProgressBar风格设置
Android系统自带的ProgressBar风格不是很好,如果想自己设置风格的话,一般有几种方法.首先介绍一下第一种方法通过动画实现.在res的anim下创建动画资源loading.xml: < ...
- 【kate总结】matlab调用opencv总结
正常情况下,编写好matlab调用opencv的代码. 1.输入 MEX XX.CPP(所有的mex都要编译) 2.将生成的.mexw64 放到要调用的文件夹下即可 出错总结: 本人写的matla ...
- @ModelAttribute注解的作用
@ModelAttribute注解的作用:1.放在方法上注解不带属性: 方法无返回值: 执行其他方法时,先执行该注解标记方法. 如果方法中有将一些属性放入model的操作,其他方法model中也会共享 ...
- poj1181 大数分解
//Accepted 164 KB 422 ms //类似poj2429 大数分解 #include <cstdio> #include <cstring> #include ...
- Canopy使用教程 (3)
1. 2. plot函数: plot默认生成是曲线图,可以通过kind参数生成其他的图形,可选的值为:line, bar, barh, kde, density, scatter. 散点图.使用kin ...
- 如何修改svn的密码或重新输入用户名密码
在Eclipse 使用SVN 的过程中大多数人往往习惯把访问SVN 的用户名密码自动保存起来以便下次自动使用,不要再次手工输入,而此时(自动保存密码后),svn又不存在一个显式的登陆框了,但是有些时候 ...
- Python学习路程day3
set集合 set是一个无序且不重复的元素集合,访问速度快,天生解决重复问题 s1 = set() s1.add('luo') s2 = set (['luo','wu','wei','ling' ...
- ios--NSCalendar NSDateComponents
原文: ios时间那点事--NSCalendar NSDateComponents http://my.oschina.net/yongbin45/blog/156181 目录[-] iOS时间那点事 ...