cf581C Developing Skills
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.
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.
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.
2 4
7 9
2
3 8
17 15 19
5
2 2
99 100
20
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
- 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 ...
- 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 ...
- 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 ...
- Developing Skills
题目传送门:点击打开链接 #include <iostream> #include <cstdio> #include <cstdlib> #include < ...
- 【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 ...
- 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 ...
- 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 ...
- 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 ...
- Codeforces Round #322 (Div. 2)
水 A - Vasya the Hipster /************************************************ * Author :Running_Time * C ...
随机推荐
- 适配ios7
if ([self respondsToSelector:@selector(edgesForExtendedLayout)]) { self.edgesForExtendedLayout = UIR ...
- Android四大组件——Activity
Activity作为Android四大组件之一,也是其中最重要的一个组件.作为一个与用户交互的组件,我们可以把Activity比较成为windows系统上的一个文件夹窗口,是一个与用户交互的界面.再进 ...
- [转] JS nodeType返回类型
将HTML DOM中几个容易常用的属性做下记录: nodeName.nodeValue 以及 nodeType 包含有关于节点的信息. nodeName 属性含有某个节点的名称. 元素节点的 node ...
- HDU -2524 矩形A + B
找规律题,这种题目比较巧妙,要仔细观察找出规律 1. 假设只有一行,一共有n列,那么由一个小矩形构成的矩形个数为n, 由两个小矩形构成的矩形个数为 n - 1个 .... 由 n 个小矩形构成的矩形个 ...
- 将用户信息保存到Cookie中
/** * 把用户保存到Cookie * * @param request * @param response * @param member */ private void rememberPwdA ...
- 更加详细的Log4net的配置
请转到周金桥的文章 http://blog.csdn.net/zhoufoxcn/article/details/6029021
- tomcat加载不了spring-webjar终极解决办法
Problems: I included: all Spring libs, Apache Tomcat 7.0 library in Build Path but it still gives er ...
- (转)SVN源码管理(上&下)
原文地址:http://www.cnblogs.com/IPrograming/archive/2012/12/15/SVN_1.html 使用SVN进行源码管理(上) 在原来的项目中使用的源码管理工 ...
- pointer-events属性
pointer-events的风格更像JavaScript,它能够: 1.阻止用户的点击动作产生任何效果.阻止缺省鼠标指针的显示3.阻止CSS里的hover和active状态的变化触发事件4.阻止Ja ...
- javascript时钟
<script language="JavaScript" type="text/javascript"> function now ...