Codeforces Round #417 (Div. 2) C. Sagheer and Nubian Market

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 contains n different items numbered from 1 to n. The i-th item has base cost a_i Egyptian pounds. If Sagheer buys k items with indices x₁, x₂, ..., x_k, then the cost of item x_j is a_xj + x_j·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, ..., 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 these k souvenirs.

Examples

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 [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 him 8 pounds, so he cannot buy it.

分析：  二分法，不断的缩小可能的区间。

要注意的是 ，物品序号是最初的序号，与我们选取的顺序无关。

每次使用二分的时候都需要重新排一次序 找到最优的取法（最优取法随着取的数量会不断变化，所以不能只排一次序）

复杂度为n（log²n）

代码如下：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
using namespace std;
typedef long long ll;
ll sum[100100];
ll sum2[100100];
ll sum3[100100];
 ll n,s,l,r,mid,ans,maxx;
ll check(ll x)
{
  if(sum2[x]<=s)
    return 1;
  return 0;
}
struct node{
    ll id;
    ll pri;
}a[100100],now[100100];
int cmp(node x,node y)
{
    return x.pri<y.pri;
}
int main()
{
    while(scanf("%lld%lld",&n,&s)!=EOF)
    {
        maxx=0;
        memset(sum,0,sizeof(sum));
        memset(sum2,0,sizeof(sum2));
        for(int i=1;i<=n;i++){
            scanf("%lld",&a[i].pri);
            a[i].id=i;
        }
        sort(a+1,a+n+1,cmp);
        for(int i=1;i<=n;i++)
        {
            sum[i]=sum[i-1]+a[i].pri;
        }
        l=0;
        r=n+1;
       while(l+1<r)
       {
         memset(sum2,0,sizeof(sum2));
         mid=(l+r)/2;
          for(int i=1;i<=n;i++)
            now[i].pri=a[i].id*mid+a[i].pri;
            sort(now+1,now+n+1,cmp);
          for(int i=1;i<=n;i++)
          {
           sum2[i]=sum2[i-1]+now[i].pri;
          }
         if(check(mid))
         {
          l=mid;
          ans=sum2[mid];
         }
         else r=mid;
       }
       printf("%lld %lld
",l,ans);
    }
    return 0;
}

　　

CF中自己的失误：没有正确理解题意，最后在A被hack之后 乱了节奏，以及自己对复杂度的错误估计放弃了在二分里面写循环。