Codeforces 812C Sagheer and Nubian Market【二分】水题

C. Sagheer and Nubian Market

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 containsn different items numbered from 1 to n. The i-th item has base cost a_i Egyptian pounds. If Sagheer buysk items with indices x₁, x₂, ..., x_k, then the cost of itemx_j isa_xj + x_j·k for1 ≤ j ≤ k. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factork.

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 andS (1 ≤ n ≤ 10⁵ and1 ≤ S ≤ 10⁹) — the number of souvenirs in the market and Sagheer's budget.

The second line contains n space-separated integersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁵) — the base costs of the souvenirs.

Output

On a single line, print two integers k,T — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy thesek souvenirs.

Examples

Input

3 112 3 5

Output

2 11

Input

4 1001 2 5 6

Output

4 54

Input

1 77

Output

0 0

Note

In the first example, he cannot take the three items because they will cost him[5, 9, 14] with total cost 28. If he decides to take only two items, then the costs will be[4, 7, 11]. So he can afford the first and second items.

In the second example, he can buy all items as they will cost him [5, 10, 17, 22].

In the third example, there is only one souvenir in the market which will cost him8 pounds, so he cannot buy it.

题目大意：

一共N个物品，你现在有S个单位的钱。

我们现在希望买的物品尽可能的多，问最多可以买多少个，用最少的花费是多少？

一个物品的价值是其本身的价值+买的物品个数*下标。

思路：

二分买的个数即可。

然后将所有物品按照当前购买量的价值从小到大排序，维护一个价值和和s比较大小判断即可。

注意上下界和数据范围。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;#define ll __int64struct node{    ll val,pos,b;}a[1500000];ll n,s,sum;ll cmp(node a,node b){    return (a.val+a.pos*a.b)<(b.val+b.pos*b.b);}ll Slove(ll mid){    for(ll i=1;i<=n;i++)a[i].b=mid;    sort(a+1,a+1+n,cmp);    sum=0;    for(ll i=1;i<=mid;i++)    {        sum+=a[i].val+a[i].pos*a[i].b;        if(sum>s)return 0;    }    if(sum<=s)return 1;    else return 0;}int main(){    while(~scanf("%I64d%I64d",&n,&s))    {        for(ll i=1;i<=n;i++)scanf("%I64d",&a[i].val),a[i].pos=i;        ll l=0;        ll r=n;        ll ans=-1;        ll ans2=-1;        while(r-l>=0)        {            ll mid=(l+r)/2;            if(Slove(mid)==1)            {                ans=mid;                ans2=sum;                l=mid+1;            }            else r=mid-1;        }        printf("%I64d %I64d\n",ans,ans2);    }}