Codeforces Round #322 (Div. 2) A、B、C

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 .

题目大意：

输入一个数，n和K，然后有n个数arr[i],让你求，在满足每个数<=100和k>0的情况下，每个值处以10取整的最大值，

注意arr[i]可以加不大于k的数，但是加完之后的值必须<=100，

解题思路：

就是先将每个值除以10之后取整的值记下来，然后再将它们对10取余之后的值记下来，再根据题意模拟判断就ok了，

但是一定要注意一些情况，arr[i]的值一直<=100

上代码：

/**
Author: ITAK

Date: 2015-09-29
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;

#define MM(a) memset(a,0,sizeof(a))

typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 100000+5;
const int mod = 1000000007;
const double eps = 1e-7;

int cmp(int a, int b)
{
    return a > b;
}
int arr[maxn], data[maxn];
int main()
{
    int n, k;
    while(~scanf("%d%d",&n,&k))
    {
        int sum=0,K=0;
        MM(data);
        for(int i=0; i<n; i++)
        {
            scanf("%d",&arr[i]);
            if(arr[i] != 100)
            {
                data[K++] = arr[i]%10;
                sum += arr[i]/10;
            }
            else
                sum += 10;
        }
        sort(data, data+K, cmp);
        for(int i=0; i<K; i++)
        {
            k -= (10-data[i]);
            data[i] =10;
            if(k<0)
                break;
            if(data[i]%10==0)
                sum ++;
        }
        if(n*10-sum > k/10)
            sum += k/10;
        else
            sum = n*10;
        cout<<sum<<endl;
    }
    return 0;
}