解题报告 之 CodeForces 581C Developing Skills

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 a_i from 0 to 100. The higher the number a_i 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 xdown 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 a_i (0 ≤ a_i ≤ 100), where a_i 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

Input

2 47 9

Output

2

Input

3 817 15 19

Output

5

Input

2 299 100

Output

20

Hint

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 lfloorfrac{100}{10} rfloor +  lfloorfrac{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 .

Source

Codeforces Round #322 (Div. 2)

题目大意：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：一道超水的题。。。还是做的太急了，又扣分了。哭。做之前先想清楚，很明显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。（注意不一定用的完，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。

上代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN = 1e5 + 10;struct Node{int num;int dig;bool operator<( const Node& rhs )const{if(dig == 10) return false;if(rhs.dig == 10)return true;return dig > rhs.dig;}};Node node[MAXN];int main(){int n, k;while(scanf( "%d%d", &n, &k ) == 2){memset( node, 0, sizeof node );int ans = 0;for(int i = 1; i <= n; i++){scanf( "%d", &node[i].num );node[i].dig = (node[i].num == 100 ? 10 : node[i].num % 10);ans += node[i].num / 10;}sort( node + 1, node + n + 1 );for(int i = 1; node[i].dig!=10&&i<=n; i++){k -= (10 - node[i].dig);if(k < 0) goto here;ans++;node[i].num += (10 - node[i].dig);node[i].dig = (node[i].num == 100 ? 10 : node[i].num % 10);}for(int i = 1; i <= n; i++){k -= 100 - node[i].num;if(k < 0){ans += (k + 100 - node[i].num) / 10;break;}ans += (100 - node[i].num) / 10;node[i].num = 100;}here:;printf( "%d\n", ans );}return 0;}