Codeforces Round #280 (Div. 2) A , B , C
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:能堆几层,一层(1-n)的和;
思路:水;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=2e5+,M=4e6+,inf=1e9+,mod=1e9+;
int a[N];
int main()
{
int x,sum=;
for(int i=;i<=;i++)
{
sum+=i;
a[i]=sum+a[i-];
}
scanf("%d",&x);
printf("%d\n",upper_bound(a+,a+,x)-(a+));
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:n个灯,L长度的路,求灯最少的照射长度,使得灯把这路全部照亮;
思路:拍个序,前后端点特判;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=2e5+,M=4e6+,inf=1e9+,mod=1e9+;
double a[N];
int main()
{
int n,l;
scanf("%d%d",&n,&l);
for(int i=;i<=n;i++)
scanf("%lf",&a[i]);
sort(a+,a++n);
double ans=;
for(int i=;i<=n;i++)
ans=max(ans,(a[i]-a[i-])/);
ans=max(a[],max(ans,l-a[n]));
printf("%f\n",ans);
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.
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=2e5+,M=4e6+,inf=1e9+,mod=1e9+;
struct grade
{
ll a,b;
bool operator <(const grade &a)const
{
return b<a.b;
}
}a[N];
int main()
{
ll n,r,ave;
scanf("%lld%lld%lld",&n,&r,&ave);
ll sum=ave*n,ans=;
for(int i=;i<=n;i++)
scanf("%lld%lld",&a[i].a,&a[i].b),sum-=a[i].a;
sort(a+,a+n+);
if(sum<=)
return puts("");
for(int i=;i<=n;i++)
ans+=min(sum,r-a[i].a)*a[i].b,sum-=min(sum,r-a[i].a);
printf("%lld\n",ans);
return ;
}
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:

Codeforces Round #280 (Div. 2) A , B , C的更多相关文章
- 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 B C 暴力 水 贪心
A. Vanya and Cubes time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- 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 ...
- Codeforces Round #280 (Div. 2)_C. Vanya and Exams
C. Vanya and Exams time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #280 (Div. 2) E. Vanya and Field 思维题
E. Vanya and Field time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #280 (Div. 2) D. Vanya and Computer Game 预处理
D. Vanya and Computer Game time limit per test 2 seconds memory limit per test 256 megabytes input s ...
- 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 ...
随机推荐
- Codeforces 678E Another Sith Tournament 状压DP
题意: 有\(n(n \leq 18)\)个人打擂台赛,编号从\(1\)到\(n\),主角是\(1\)号. 一开始主角先选一个擂主,和一个打擂的人. 两个人之中胜的人留下来当擂主等主角决定下一个人打擂 ...
- android菜鸟学习笔记31----Android使用百度地图API(二)获取地理位置及地图控制器的简单使用
1.获取当前地理位置: Android中提供了一个LocationManager的类,用于管理地理位置.不能通过构造函数获取该类的实例,而是通过Context的getSystemService(): ...
- 巨蟒python全栈开发linux之centos7
1.crm项目部署回顾(小BOSS) crm部署 nginx+uwsgi+django+mysql nginx 前端 uwsgi+django 后端 mysql 数据支撑 crm是一 ...
- General Decimal Arithmetic 浮点算法
General Decimal Arithmetic http://speleotrove.com/decimal/ General Decimal Arithmetic [ FAQ | Decima ...
- 【python】-- Django 分页 、cookie、Session、CSRF
Django 分页 .cookie.Session.CSRF 一.分页 分页功能在每个网站都是必要的,下面主要介绍两种分页方式: 1.Django内置分页 from django.shortcuts ...
- 自定义admin
平时我们用的django自带admin,怎么评价呢?一个字简陋,而且也人性化,如下图,首先只显示数据对象,如果要查看详细还有点进去,其次不能对自己想要的数据进行刷选 我们的期望是:数据如excel显示 ...
- python作用域和JavaScript作用域
JavaScript 一.JavaScript中无块级作用域 一个大括号一个作用域,就属于块级作用域,在Java和c#才存在块级作用域 function Main(){ if(1==1){ var n ...
- jquery生成二维码图片
1.插件介绍 jquery.qrcode.min.js插件是jq系列的基于jq,在引入该插件之前要先引入jq.能够在客户端生成矩阵二维码QRCode 的jquery插件 ,使用它可以很方便的在页面上生 ...
- python并发编程&多线程(二)
前导理论知识见:python并发编程&多线程(一) 一 threading模块介绍 multiprocess模块的完全模仿了threading模块的接口,二者在使用层面,有很大的相似性 官网链 ...
- Symfony 一些介绍
Symfony 一些介绍: 路由:能限制 hostname,这就让有大量公共功能的网站可以共用一套代码:URI 识别支持 Reg 检测,让 url 能定义的随心所欲:支持前缀,import,便于模块化 ...