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. appium获取toast方法

    配置toast请注意: 1.指定desired_caps["automationName"] = "UiAutomator2" 2.要求安装jdk1.8 64位 ...

  2. Redis报错:redis.exceptions.ResponseError: MISCONF Redis is configured to save RDB snap

    首先找到出现错误的原因: redis.exceptions.ResponseError: MISCONF Redis is configured to save RDB snapshots, but ...

  3. (转)jieba中文分词的.NET版本:jieba.NET

    简介 平时经常用Python写些小程序.在做文本分析相关的事情时免不了进行中文分词,于是就遇到了用Python实现的结巴中文分词.jieba使用起来非常简单,同时分词的结果也令人印象深刻,有兴趣的可以 ...

  4. Java异常机制关键字总结,及throws 和 throw 的区别

    在Java的异常机制中,时常出现五个关键字:try , catch , throw , throws , finally. 下面将总结各个关键字的用法,以及throw和throws的区别: (1) t ...

  5. JS检测数据类型

    如果你要判断的是基本数据类型或JavaScript内置对象,使用toString: 如果要判断的时自定义类型,请使用instanceof. 1.typeof typeof操作符返回的是类型字符串,它的 ...

  6. 从setTimeout谈js运行机制

    众所周知,JavaScript是单线程的编程,什么是单线程,就是说同一时间JavaScript只能执行一段代码,如果这段代码要执行很长时间,那么之后的代码只能尽情地等待它执行完才能有机会执行,不像人一 ...

  7. eclipse中提示js或者JQuery代码

    当你在eclipse中的JSP中写JavaScript或者JQuery代码的时候,eclipse是不会自动提示的,所以你需要在eclipse中安装一下插件,该插件的名字叫:Spket IDE,它可以作 ...

  8. Linux下SSH工具 PAC Manager的安装

    PAC Manager, Linux下类似SecureCRT Xshell的SSH工具,该工具功能上相当的不错,完全可以代替SecureCRT Xshell的功能. PAC (Perl Auto Co ...

  9. A GDB Tutorial with Examples--转

    http://www.cprogramming.com/gdb.html A GDB Tutorial with Examples By Manasij Mukherjee A good debugg ...

  10. 多功能电子通讯录(涉及到了双向链表的使用,Linux文件编程等等)

    readme.txt //作为一个程序员,我们咋么能不写用户手册呢!MSP的我觉得用户体验是王道,苹果手机的用户体验的确不错!不过WP加油!我去,扯远了!赶紧看我的程序吧!  歡迎使用多功能電子通訊錄 ...