20190216_E

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

题意：一开始有n个技能，k个技能点。任何一个技能达达到10级，人物就升1级，技能最高100级。求任务最高能升到多少级？

思路：
1、先算人物的初始等级。
2、用10 - （a[i] mod 10）算出它们各个技能离10级还差多少技能点，然后sort
3、累加技能点（c）大于k时，结束循环
4、判断是否有剩余技能点，如果k-c大于10，则继续升级，但最高等级不能超过n*10.
下面是代码：：：：

#include<stdio.h>
#include<iostream>
#include<map>
#include<algorithm>
#include<cstring>
#include<string.h>
#include<math.h>
using namespace std ;
typedef long long ll;
#define maxn 10000005
#define INF 0x3f3f3f3f//将近int类型最大数的一半，而且乘2不会爆int
int main( )
{
    int n, k, a[100005], sum = 0, c=0, s=0;
    cin >> n >> k;
    for(int i=0; i<n; ++i)
    {
        scanf("%d", &a[i]);
        sum+=(a[i]/10);
        if(a[i]%10!=0)
            a[i] = (10-a[i]%10);
        else
            a[i] = 0;
    }
    sort(a, a+n);
     for(int i=0; i<n; ++i)
    {
        if(a[i] == 0)
            continue;
        else
        {
            c+=a[i];
            if(c>k)
                break;
            else
                s++;
        }
    }
    if(k>c)
        s+=(k-c)/10;
    sum+=s;
    if(sum > n*10)
        sum=n*10;
    cout << sum << '\n';
    return 0;
}

就是这样子了。。。。