贪心，选出前m个最大翻转可增加的值

解题思路:

注意事项:

参考代码:

#include <iostream>

#include <algorithm>

#include <cstring>

#include <vector>

#include <cmath>

using namespace std;

typedef long long ll;

const int N = 1010;

int n,m;

ll a[N],f[N];

ll ans = 0;

vector<ll> sum;

ll rev(int num)   //翻转当前数

{

vector<int> w;

ll s = 0;

while(num)

{

     if(num%2 == 0) w.push_back(0);

     else w.push_back(1);

     num/=2;

} 

for(int i=0;i<w.size();i++)

{

    if(w[i] == 1) s+=pow(2,w.size()-i-1);

}

return s;

}

bool cmp(int a,int b) {return a > b;}   //从大到小排序

int main()

{

     cin >> n >> m;

     for(int i=0;i<n;i++) 

     {

         cin >> a[i];

         ans+=a[i];

         f[i] = rev(a[i]);

     }

     ll t = 0;

     for(int i=0;i<n;i++)  

     {

         if(f[i] > a[i]) 

         {

              t+=f[i]-a[i];

         }

         else

         {

              sum.push_back(t);

              t = 0;

         }

     }

     if(t > 0) sum.push_back(t);

     sort(sum.begin(),sum.end(),cmp);

     for(int i=0;i<m;i++)

     {

          ans+=sum[i];

     }

     cout << ans;

     return 0;

}