Codeforces Round #322 (Div. 2) C. Developing Skills 优先队列

C. Developing Skills

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/581/problem/C

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.

Sample Input

2 4
7 9

Sample Output

2

HINT

题意

有一个人有n个技能，有k个技能点

每个技能最高10级，每点十个技能点算一级

然后问你最后级数之和最高为多少

题解：

优先队列贪心就好了

优先加离10进的数

代码：

//qscqesze
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <bitset>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1205000
#define mod 1000000007
#define eps 1e-9
#define e exp(1.0)
#define PI acos(-1)
#define lowbit(x) (x)&(-x)
const double EP  = 1E-10 ;
int Num;
//const int inf=0x7fffffff;
const ll inf=999999999;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//*************************************************************************************

struct node
{
    int x;
    bool operator<(const node& other)  const
    {
        if((x%10)==(other.x%10))return x>other.x;
        return (x%10)<(other.x%10);
    }
};

priority_queue<node> Q;
int main()
{
    int n=read(),k=read();
    for(int i=0;i<n;i++)
    {
        int x=read();
        Q.push(node{x});
    }
    while(k)
    {
        node now = Q.top();
        if(now.x==100)break;
        int p = min(10-now.x%10,k);
        now.x+=p;
        k-=p;
        Q.pop();
        Q.push(now);
    }
    int ans = 0;
    for(int i=0;i<n;i++)
    {
        node t = Q.top();
        ans += t.x/10;
        Q.pop();
    }
    cout<<ans<<endl;
}