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 numberai 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 test(s)
input
2 4
7 9
output
2
input
3 8
17 15 19
output
5
input
2 2
99 100
output
20
Note

In the first test case the optimal strategy is as follows. Petya has to improve the first skill to 10 by spending 3 improvement units, and the second skill to 10, by spending one improvement unit. Thus, Petya spends all his improvement units and the total rating of the character becomes equal to lfloor frac{100}{10} rfloor +  lfloor frac{100}{10} rfloor = 10 + 10 =  20.

In the second test the optimal strategy for Petya is to improve the first skill to 20 (by spending 3 improvement units) and to improve the third skill to 20 (in this case by spending 1 improvement units). Thus, Petya is left with 4 improvement units and he will be able to increase the second skill to 19 (which does not change the overall rating, so Petya does not necessarily have to do it). Therefore, the highest possible total rating in this example is .

In the third test case the optimal strategy for Petya is to increase the first skill to 100 by spending 1 improvement unit. Thereafter, both skills of the character will be equal to 100, so Petya will not be able to spend the remaining improvement unit. So the answer is equal to .

题意是n个技能加点,有m点可以加,每个技能点数不能超过100。最后战斗力是Σ(a[i]/10)(是整除10)。求最大战斗力。

这贪心搞一下最后一位然后随便做就好了。。

一开始没处理全100的时候依然往上加结果gg……wa来wa去一时爽

 #include<set>
#include<map>
#include<cmath>
#include<ctime>
#include<deque>
#include<queue>
#include<bitset>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
#define inf 0x7fffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int n,m,ans;
int a[];
inline bool cmp(int a,int b)
{
int x=a%,y=b%;x=-x;y=-y;
if (x<y)return ;
if (y<x)return ;
return a<b;
}
int main()
{
n=read(),m=read();
for(int i=;i<=n;i++)a[i]=read(),ans+=a[i]/;
sort(a+,a+n+,cmp);
for (int i=;i<=n;i++)
{
if (a[i]==)continue;
int x=a[i]%;x=-x;
if(x>m)break;
m-=x;a[i]+=x;ans++;
}
sort(a+,a+n+);
for (int i=;i<=n;i++)while (a[i]<=&&m>=)ans++,a[i]+=,m-=;
printf("%d\n",ans);
}

cf581C

cf581C Developing Skills的更多相关文章

  1. CF581C Developing Skills 模拟

    Petya loves computer games. Finally a game that he's been waiting for so long came out! The main cha ...

  2. 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/p ...

  3. codeforces 581C. Developing Skills 解题报告

    题目链接:http://codeforces.com/problemset/problem/581/C 题目意思:给出 n 个数:a1, a2, ..., an (0 ≤ ai ≤ 100).给出值 ...

  4. Developing Skills

    题目传送门:点击打开链接 #include <iostream> #include <cstdio> #include <cstdlib> #include < ...

  5. 【Henu ACM Round#19 C】 Developing Skills

    [链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 优先把不是10的倍数的变成10的倍数. (优先%10比较大的数字增加 如果k还有剩余. 剩下的数字都是10的倍数了. 那么先加哪一个 ...

  6. codeforces581C

    Developing Skills CodeForces - 581C 你在玩一个游戏.你操作的角色有n个技能,每个技能都有一个等级ai.现在你有k次提升技能的机会(将其中某个技能提升一个等级,可以重 ...

  7. How do I learn mathematics for machine learning?

    https://www.quora.com/How-do-I-learn-mathematics-for-machine-learning   How do I learn mathematics f ...

  8. 每日英语:A Better Way To Treat Anxiety

    Getting up the nerve to order in a coffee shop used to be difficult for 16-year-old Georgiann Steely ...

  9. Codeforces Round #322 (Div. 2)

    水 A - Vasya the Hipster /************************************************ * Author :Running_Time * C ...

随机推荐

  1. hdu 4540 威威猫系列故事——打地鼠 dp小水题

    威威猫系列故事——打地鼠 Time Limit: 300/100 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total ...

  2. 一起来写2048(160行python代码)

    前言: Life is short ,you need python. --Bruce Eckel 我与2048的缘,不是缘于一个玩家,而是一次,一次,重新的ACM比赛.四月份校赛初赛,第一次碰到20 ...

  3. 安装LVS安装LVS和配置LVS的工作比较繁杂

    安装LVS安装LVS和配置LVS的工作比较繁杂,读者在配置的过程中需要非常细心和耐心.在本节我们将对其进行详细地介绍.主要包括如下几个核心步骤:1.获取支持LVS的内核源代码如果读者需要使用LVS,需 ...

  4. 阅读underscore源码笔记

    本文为原创作品,可以转载,但请添加本文连接,谢谢传阅,本人博客已转移至github,地址为:jruif.github.io underscorejs,一个实用的的Javascript函数库,值得推荐, ...

  5. Blade和其他构建工具有什么不同

    大部分人都至少接触过不止一种构建工具,比如make,autotools.而我们开发了Blade,为什么那么多现成的工具不用,而又再造了一个轮子,相对于传统的make等工具,Blade的好处在又哪里呢? ...

  6. Linux LVM 扩展磁盘分区

    系统:centos 6.3--新建分区 fdisk -l /dev/sdc       # 查看分区 fdisk /dev/sdc          # 创建分区 :n                 ...

  7. hibernate注解原理

    持续更新中.. hibernate注解用的是java注解,用到的是java反射机制.

  8. CSS3 过滤

    CSS3 过滤 通过CSS3,我们可以在不适用flash动画或JavaScript的情况下,当元素从一种样式变换为另一种样式时为元素添加效果. 浏览器支持 属性 浏览器支持 transition   ...

  9. (转)jQuery LigerUI 插件介绍及使用之ligerTree

    一,简介  ligerTree的功能列表: 1,支持本地数据和服务器数据(配置data或者url) 2,支持原生html生成Tree 3,支持动态获取增加/修改/删除节点 4,支持大部分常见的事件 5 ...

  10. mySQL优化 my.ini 配置说明

    [mysqld] port = 3306 serverid = 1 socket = /tmp/mysql.sock skip-name-resolve #禁止MySQL对外部连接进行DNS解析,使用 ...