Codeforces C. Duff and Weight Lifting

最新推荐文章于 2019-08-21 21:33:00 发布

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本文介绍了一种算法问题——如何最小化完成举重任务所需的步骤。通过将重量表示为二的幂次形式，利用这些权重值进行组合，每次组合的目标是形成一个新的二的幂次值。文章详细阐述了解决方案的思路与实现代码，关键在于统计每个数字出现的次数，并通过迭代合并来减少操作步骤。

C. Duff and Weight Lifting
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

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.
Examples
Input

5
1 1 2 3 3

Output

Input

4
0 1 2 3

Output

Note

In the first sample case: One optimal way would be to throw away the first three in the first step and the rest in the second step. Also, it’s not possible to do it in one step because their sum is not a power of two.

In the second sample case: The only optimal way is to throw away one weight in each step. It’s not possible to do it in less than 4 steps because there’s no subset of weights with more than one weight and sum equal to a power of two.

大致思想：两个低位可以组成一个高位，所以我们可以统计每一个数字的个数，然后再从0到N枚举即可，是奇数则多一步，而偶数合并到最后也一定会是奇数
代码如下：

#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define inf 0x3f3f3f3f
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rep1(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=1e6+100;
int arr[N];
int vis[N];
int main()
{
    ios::sync_with_stdio(false);
    int n;
    cin>>n;
    map<int,int>mp  ;
    memset(vis,0,sizeof vis);
    for(int i=0;i<n;i++)
    {
        cin>>arr[i];
        vis[arr[i]]++;
    }
    sort(arr,arr+n);
    int cnt=0;
    for(int i=0;i<N;i++)
    {
        vis[i+1]+=vis[i]/2;
        if(vis[i]%2!=0)
            cnt++;
    }
    cout<<cnt<<endl;
    return 0;
}