The Little Girl who Picks Mushrooms HDU

当n<=3是肯定可以带回去1024因为可以用另外一包跟n包凑成给仙女的 然后另外一包或两包是1024

当n=4时 两种情况 1.选已采集的两包蘑菇，跟待采集的1包凑成1024的整数倍，让剩下的最多

已存在3包能凑成整数倍，待采集的那一包跟已采集的凑成1024

N=5时判断有没有3包组成1024的整数倍即可。

但是n为4或5时有个很坑的地方就是可以给仙女0个。

#include <iostream>

#include<cstdio>

#include<algorithm>

using namespace std;

 

int main()

{

    int n;

    while(cin>>n)

    {

        int a[5];

        for(int i=0;i<n;i++)

        {

            scanf("%d",&a[i]);

        }

        if(n<=3)cout<<"1024"<<endl;

        else if(n==4)

        {

            //sort(a,a+4);

            int mmax=-999999;

            int sum=a[0]+a[1]+a[2]+a[3];

            for(int i=0;i<4;i++)

            {

                for(int j=0;j<4;j++)

                {

                    if(i!=j)

                    {

                        mmax=max(mmax,(sum-a[i]-a[j])%1024);

                        if((sum-a[i]-a[j])%1024==0&&(sum-a[i]-a[j])!=0)

                                mmax=1024;

                    }

 

                }

            }

             for(int i=0;i<4;i++)

            {

                for(int j=0;j<4;j++)

                {

                    for(int k=0;k<4;k++)

                    {

                        if((i!=j)&&(j!=k)&&(i!=k))

                        {

                            if((a[i]+a[j]+a[k])%1024==0)

                            {

                                mmax=1024;

                            }

                        }

                    }

                }

            }

            if(mmax<0)mmax=0;

            cout<<mmax<<endl;

        }

        else

        {

            //sort(a,a+5);

            int sum=a[0]+a[1]+a[2]+a[3]+a[4];

            int mmax=-999999;

            for(int i=0;i<5;i++)

            {

                for(int j=0;j<5;j++)

                {

                    for(int k=0;k<5;k++)

                    {

                        if((i!=j)&&(j!=k)&&(i!=k))

                        {

                            if((a[i]+a[j]+a[k])%1024==0)

                            {

                                mmax=max(mmax,(sum-a[i]-a[j]-a[k])%1024);

                                if((sum-a[i]-a[j]-a[k])%1024==0&&(sum-a[i]-a[j]-a[k])!=0)

                                mmax=1024;

                            }

                        }

                    }

                }

            }

            if(mmax<0)mmax=0;

            cout<<mmax<<endl;

        }

    }

    return 0;

}