time limit per test  2 seconds
memory limit per test  256 megabytes
 

On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains n different items numbered from 1 to n. The i-th item has base cost ai Egyptian pounds. If Sagheer buysk items with indices x1, x2, ..., xk, then the cost of item xj is axj + xj·k for 1 ≤ j ≤ k. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor k.

Sagheer wants to buy as many souvenirs as possible without paying more than S Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?

Input

The first line contains two integers n and S (1 ≤ n ≤ 105 and 1 ≤ S ≤ 109) — the number of souvenirs in the market and Sagheer's budget.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the base costs of the souvenirs.

Output

On a single line, print two integers kT — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these ksouvenirs.

 
input
3 11
2 3 5
output
2 11
input
4 100
1 2 5 6
output
4 54
input
1 7
7
output
0 0
Note
In the first example, he cannot take the three items because they will cost him [, , ] with total cost . If he decides to take only two items, then the costs will be [, , ]. So he can afford the first and second items. In the second example, he can buy all items as they will cost him [, , , ]. In the third example, there is only one souvenir in the market which will cost him pounds, so he cannot buy it.

Note


题目大意:
     在一个商店里一共有n件商品,每一件都有一个基础价格,但是这个商店有一个奇怪的规定,
     每一件商品最后的价格为 基础价格+买的商品总件数*商品的坐标(即这是第几件商品)
     现给定 商品的件数和你所拥有的钱数, 问你最多能卖几件商品 花的总钱数是多少 解题思路:
     总体思路为:二分查找 先找出最多能买的商品的件数,然后在计算出总花费 AC代码:
 #include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll; int main ()
{
ll n,s,sum,x,y,l,r;
ll ans[],a[];
int i;
cin>>n>>s;
for (i = ; i <= n; i ++)
cin>>a[i]; l = ,r = n; // 左标签和右标签
x = y = ;
while (l <= r){
sum = ;
ll mid = (r+l)>>; // 二分枚举能购买的商品件数
for (i = ; i <= n; i ++)
ans[i] = a[i]+i*mid;
sort(ans+,ans++n);
for (i = ; i <= mid; i ++)
sum += ans[i];
if (sum <= s){
x = mid;
y = sum;
l = mid+;
}
else
r = mid-;
}
cout<<x<<" "<<y<<endl;
return ;
}
以下版本思路同上;
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; int main ()
{
__int64 n,s;
__int64 ans[],a[];
int i;
while (~scanf("%I64d%I64d",&n,&s))
{
for (i = ; i <= n; i ++)
scanf("%I64d",&a[i]); int l = ,r = n;
__int64 sum,x=,y=;
while (l <= r){
sum = ;
long long mid = (l+r)>>;
for (i = ; i <= n; i ++)
ans[i] = a[i]+i*mid;
sort(ans+,ans++n);
for (i = ; i <= mid; i ++)
sum += ans[i];
if (sum > s)
r = mid-;
else
l = mid+;
}
for (i = ; i <= n; i ++)
ans[i] = a[i]+i*r;
sort(ans+,ans++n);
for (i = ; i <= r; i ++)
y += ans[i];
printf("%d %I64d\n",r,y);
}
return ;
}

 

Codeforces Round #417 C. Sagheer and Nubian Market的更多相关文章

  1. Codeforces Round #417 B. Sagheer, the Hausmeister

    B. Sagheer, the Hausmeister time limit per test  1 second memory limit per test  256 megabytes   Som ...

  2. AC日记——Sagheer and Nubian Market codeforces 812c

    C - Sagheer and Nubian Market 思路: 二分: 代码: #include <bits/stdc++.h> using namespace std; #defin ...

  3. Codeforces J. Sagheer and Nubian Market(二分枚举)

    题目描述: Sagheer and Nubian Market time limit per test 2 seconds memory limit per test 256 megabytes in ...

  4. CodeForce-812C Sagheer and Nubian Market(二分)

    Sagheer and Nubian Market CodeForces - 812C 题意:n个货物,每个货物基础价格是ai. 当你一共购买k个货物时,每个货物的价格为a[i]+k*i. 每个货物只 ...

  5. [Codeforces Round#417 Div.2]

    来自FallDream的博客,未经允许,请勿转载,谢谢. 有毒的一场div2 找了个1300的小号,结果B题题目看错没交  D题题目剧毒 E题差了10秒钟没交上去. 233 ------- A.Sag ...

  6. Codeforces812C Sagheer and Nubian Market 2017-06-02 20:39 153人阅读 评论(0) 收藏

    C. Sagheer and Nubian Market time limit per test 2 seconds memory limit per test 256 megabytes input ...

  7. Codeforces Round #417 (Div. 2)A B C E 模拟 枚举 二分 阶梯博弈

    A. Sagheer and Crossroads time limit per test 1 second memory limit per test 256 megabytes input sta ...

  8. Codeforces Round #417 (Div. 2) 花式被虐

    A. Sagheer and Crossroads time limit per test 1 second memory limit per test 256 megabytes input sta ...

  9. codeforces round 417 div2 补题 CF 812 A-E

    A Sagheer and Crossroads 水题略过(然而被Hack了 以后要更加谨慎) #include<bits/stdc++.h> using namespace std; i ...

随机推荐

  1. Vagrant 创建虚拟机

    Vagrant  创建虚拟机 1. 下载相关软件 虚拟机软件:vmware  virtualbox Vagrant 软件:vagrant cd /tmpwget http://download.vir ...

  2. (转)Python数学函数

    原文:https://www.cnblogs.com/lpl1/p/7793645.html PYTHON-基础-内置函数小结----------http://www.wklken.me/posts/ ...

  3. JVM-ClassLoader类加载器

    类加载器: 对于虚拟机的角度来看,只存在两种类加载器: 启动类加载器(Brootstrap ClassLoader)和“其他类加载器”.启动类加载器是由C++写的,属于虚拟机的一部分,其他类加载器都是 ...

  4. 最新 IntelliJ Idea 2017 激活方法(转)

    转载地址:http://www.cnblogs.com/suiyueqiannian/p/6754091.html 1. 到网站 http://idea.lanyus.com/ 获取注册码. 2.填入 ...

  5. Go RabbitMQ(五)主题

    RabbitMQ topic 在之前我们将交换器的类型从fanout设置为direct后能够根据我们的选择获得响应的消息,虽然改良我们的消息日志系统,但是还有很多局限性,比如它不能基于多个标准进行路由 ...

  6. 图解ARP协议(五)免费ARP:地址冲突了肿么办?

    一.免费ARP概述 网络世界纷繁复杂,除了各种黑客攻击行为对网络能造成实际破坏之外,还有一类安全问题或泛安全问题,看上去问题不大,但其实仍然可以造成极大的杀伤力.今天跟大家探讨的,也是技术原理比较简单 ...

  7. canvas的measureText()方法

    做一个小动画的时候遇到了个问题,就是在给文字应用渐变色的时候,不知怎么设置渐变色的区域. 渐变依赖于定义时的区域,超出这个区域只有纯色,而不是渐变色. 所以要取得文字的宽度. 查了资料得知,canva ...

  8. XML的基本概念和Android下的使用

    1. XML的基本概念 1. 什么是XML: 1). XML是指可扩展标记语言(eXtensible Markup Language),它是一种标记语言,很类似HTML.它被设计的宗旨是表示数据,而非 ...

  9. IntelliJ使用指南—— 导入Eclipse的Web项目

    通常一个团队中可能有人用eclipse,有人用intelliJ,那么经常会出现需要导入别人用eclipse建好的web项目.而IntelliJ提供了多种项目类型的导入方式,其中就有eclipse. 在 ...

  10. c#创建window服务

    Windows服务在Visual Studio 以前的版本中叫NT服务,在VS.net启用了新的名称.用Visual C# 创建Windows服务不是一件困难的事,本文就将指导你一步一步创建一个Win ...