cf水题 --Duff and Weight Lifting

最新推荐文章于 2020-03-02 15:18:22 发布

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最新推荐文章于 2020-03-02 15:18:22 发布

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Duff正在练习举重，她面临着一项任务，需要处理一系列重量，每个重量是2的幂。Malek要求她以最少的步骤将所有重量扔掉。在每一步，Duff可以选择一些剩余的重量，如果这些重量的总和可以表示为一个2的幂，则可以丢弃它们。问题是找到使步骤数最小化的策略。输入包含重量的数量和每个重量的2的幂值，输出应为所需的最小步骤数。

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http://codeforces.com/contest/588/my

Recently, Duff has been practicing weight lifting. As a hard practice, Malek gave her a task. He gave her a sequence of weights. Weight of i-th of them is 2wi pounds. In each step, Duff can lift some of the remaining weights and throw them away. She does this until there's no more weight left. Malek asked her to minimize the number of steps.

$\text{[math]}$

Duff is a competitive programming fan. That's why in each step, she can only lift and throw away a sequence of weights 2a1, ..., 2ak if and only if there exists a non-negative integer x such that 2a1 + 2a2 + ... + 2ak = 2x, i. e. the sum of those numbers is a power of two.

Duff is a competitive programming fan, but not a programmer. That's why she asked for your help. Help her minimize the number of steps.

Input

The first line of input contains integer n (1 ≤ n ≤ 106), the number of weights.

The second line contains n integers w1, ..., wn separated by spaces (0 ≤ wi ≤ 106 for each 1 ≤ i ≤ n), the powers of two forming the weights values.

Output

Print the minimum number of steps in a single line.

   Sample test(s) 
 
     input 
   
5
1 1 2 3 3

     output 
   
2

     input 
   
4
0 1 2 3

     output 
   
4

题目意思，给一个数n，下面有n个数wi，如果sum(2^wi)==2^x(x是整数），那么就给结果 ans 加一次，min（ans);;;

理解：

由于wi很大，无法表示2^wi,所以考虑到wi相等，k个wi=（wi+1）*k/2, k是偶数时，wi被完全合并，k是奇数时，ans++;遍历，由1---10^6+32;下面给出代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const  int maxn=1000000+32;
int a[maxn];
int main()
{
    int n,ans,data;
    while(scanf("%d",&n)!=-1)
    {

        memset(a,0,sizeof(a));
        for(int i=0;i<n;i++)
        {
            scanf("%d",&data);
            a[data]++;
        }
        ans=a[0]%2;
        for(int i=1;i<maxn;i++)
        {
            a[i]=a[i-1]/2+a[i];
            if(a[i]%2==1)
            ans++;
        }
        printf("%d\n",ans);

    }
    ///cout << "Hello world!" << endl;
    return 0;
}