CF 581C Developing Skills

C. Developing Skills

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 aiis,
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 kor
less improvement units.

Sample test(s)

input

2 4
7 9

output

2

input

3 8
17 15 19

output

5

input

2 299 100

output

20

Note

题目大意（来自网上）：有n个技能，每个技能有一个初始熟练度( [1,100] )。现在有k个技能点，每个技能点可以让某个技能的熟练度+1（可以重复给同一个技能加熟练度，但熟练度不能超过100）。最后的得分是每个技能的熟练度整除10的求和。即∑（(int)(skill/10)）。问最后得分的最大值是多少？

分析：模拟的方法值得学习，自己写不出来，知道方法后却能AC。

应该贪心先给尾数是9的加熟练度，接着给尾数为8的……这样一轮完了之后大家都以0结尾。

此时如果k还剩余，则看看还能加几个10。（这个时候再遍历一遍数组，统计出能加的最多的10的个数，应为不一定用的完，可能所有技能都已经100了）。

于是乎，先将所有的熟练度贪心平齐为0结尾，然后从1~n开始扫描，看可以再加出去多少个10，则得到最终答案。

#include<iostream>
#include<cstring>
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
pair<int,int> num[100002];
bool commp(pair<int,int> a,pair<int,int> b)
{
return a.first>b.first;
}
int main()
{
int n,m;
cin >> n >> m;
int ans = 0;
for(int i=0;i<n;i++)
{
cin >> num[i].second;
num[i].first = num[i].second%10;
ans+=num[i].second/10;
}
sort(num,num+n,commp);
for(int i=0;i<n&&m>0;i++)
{
if((10-num[i].first<=m) && (num[i].second<100))
{
ans++;
num[i].second += (10-num[i].first);
m-=(10-num[i].first);
}
else
break;
}
if(m>0)
{
int tot = 0;
for(int i=0;i<n;i++)
{
if(num[i].second<100)
tot+=(100-num[i].second)/10;
}
ans+= min(m/10,tot);
}
//ans+= m/10;

cout<<ans<<endl;
return 0;
}