学步园

200： 水题

class TaroGrid {
public:
	int getNumber(vector <string>);
};

int TaroGrid::getNumber(vector <string> grid)
{
    int ans=0;
    int black=0,white=0;
    int N=grid.size();
    for(int i=0;i<N;i++)
    {
        black=0;white=0;
        if(grid[0][i]=='B') black++;
        else white++;
        for(int j=1;j<N;j++)
        {
            if(grid[j][i]=='W')
            {
                if(black)
                {
                    ans=max(ans,black);
                    black=0;
                    white=1;
                }
                else if(white)
                {
                    white++;
                }
            }
            else if(grid[j][i]=='B')
            {
                if(black)
                {
                    black++;
                }
                else if(white)
                {
                    ans=max(ans,white);
                    white=0;
                    black=1;
                }
            }
        }
        ans=max(ans,max(white,black));
    }
    return ans;
}

500：尽量往左边靠，判断每一团猫能不能平铺

class CatsOnTheLineDiv2 {
public:
	string getAnswer(vector <int>, vector <int>, int);
};

struct CAT
{
    int p,c;
}cat[55];

bool cmp(CAT x,CAT y)
{
    if(x.p!=y.p)
        return x.p<y.p;
    else
        return x.c<y.c;
}

string CatsOnTheLineDiv2::getAnswer(vector <int> pos, vector <int> count, int time)
{
    int n=pos.size();
    for(int i=0;i<n;i++)
    {
        cat[i].p=pos[i];
        cat[i].c=count[i];
    }
    sort(cat,cat+n,cmp);
    bool flag=true;
    int left=-(1<<30),right=-(1<<30);
    for(int i=0;i<n;i++)
    {
        left=max(right+1,cat[i].p-time);
        right=left+cat[i].c-1;
        if(right-cat[i].p>time)
            flag=false;
    }
    if(flag) return "Possible";
    else return "Impossible";
}

950：记忆化搜索

map<LL,LL> ret;

class TaroCoins {
public:
    LL dfs(LL n)
    {
        if(ret.count(n)) return ret[n];
        LL ans=0;
        if(n%2LL) ans=dfs(n/2LL);
        else ans=dfs(n/2LL)+dfs(n/2LL-1LL);
        ret[n]=ans;
        return ans;
    }
	long long getNumber(long long N)
	{
	    ret[0]=1;
        return dfs(N);
	}
};