Codeforces 581C - Developing Skills(貪心)

Description

Petya loves computer games. Finally a game that he’s been waiting for so long came out!

The main character of this game has n different skills, each of which is characterized by an integer ai from 0 to 100. The higher the number ai is, the higher is the i-th skill of the character. The total rating of the character is calculated as the sum of the values ​​of for all i from 1 to n. The expression ⌊ x⌋ denotes the result of rounding the number x down to the nearest integer.

At the beginning of the game Petya got k improvement units as a bonus that he can use to increase the skills of his character and his total rating. One improvement unit can increase any skill of Petya’s character by exactly one. For example, if a4 = 46, after using one imporvement unit to this skill, it becomes equal to 47. A hero’s skill cannot rise higher more than 100. Thus, it is permissible that some of the units will remain unused.

Your task is to determine the optimal way of using the improvement units so as to maximize the overall rating of the character. It is not necessary to use all the improvement units.

Input

The first line of the input contains two positive integers n and k(1 ≤ n ≤ 105,0 ≤ k ≤ 107) — the number of skills of the character and the number of units of improvements at Petya’s disposal.

The second line of the input contains a sequence of n integers ai (0 ≤ ai ≤ 100), where ai characterizes the level of the i-th skill of the character.

Output

The first line of the output should contain a single non-negative integer — the maximum total rating of the character that Petya can get using k or less improvement units.

Examples

input

2 4
7 9

output

2

input

3 8
17 15 19

output

5

input

2 2
99 100

output

20

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Solution

題目大意：有n個技能，k個技能點。技能對應的等級爲a[i] i∈[0.n−1] ,任何一個技能，每達到10級，人物等級+1，技能最高100級。合理的加技能，求人物可以達到的最高等級是多少 ?

解題思路：

1. 先將人物初始等級統計出來。
2. b[i]=a[i] mod 10
3. 對 b[i] 進行升序排序,從高到低貪心求解。
4. k−10+b[i]>=0 ,則說明還可以繼續加技能，繼續向後執行。
5. 人物最高等級爲 n∗10 ,最終人物等級不能超過n∗10 .

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#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 11111111

int a[N], b[N];

int main( )
{
    int n, k;
    while (~scanf("%d%d", &n, &k))
    {
        int ans = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            ans += a[i] / 10;
            b[i] = a[i] % 10;
        }
        sort(b, b + n);
        for (int i = n - 1; i >= 0; i--)
        {
            if (k - 10 + b[i] >= 0)
            {
                ans++;
                k = k - 10 + b[i];
                if (k >= 10 && i == 0) ans += k / 10;
                if (ans >= 10 * n)  ans = 10 * n;
            }

            else break;
        }
        printf("%d\n", ans);
    }
    return 0;
}