2018头条春招笔试题解

原文链接：https://www.nowcoder.com/discuss/70299?type=2&order=0&pos=9&page=2

头条题目:  https://blog.csdn.net/flushhip/article/details/79685488

头条测试: https://blog.csdn.net/flushhip/article/details/79562199

第一题：

双指针：

#include
using namespace std;
typedef long long ll;
const int N = 1e6+7;
int a[N];
int main()
{
   int n,k;
   scanf ( "%d%d" ,&n,&k);
   for ( int i=0;i scanf ( "%d" ,&a[i]);
   sort(a, a+n);
   n = unique(a, a+n) -a;
   int r = 0, ans=0;
   for ( int l=0; l
   {
       while (r
       if (r==n) break ;
       if (a[r]-a[l] == k) ++ans;
   }
   printf ( "%d\n" , ans);
   return 0;
}

第二题：

BFS

#include
using namespace std;
typedef pair< int , int > pii;
map int , int > , int > mp;
int main()
{
   int n;
   scanf ( "%d" ,&n);
   queue q;
   q.push(make_pair(1, 1));
   mp[make_pair(1,1)]=0;
   while (!q.empty())
   {
       pii pr = q.front();q.pop();
//        printf("%d %d\n",pr.first, pr.second);
       if (pr.first == n)
       {
           printf ( "%d\n" , mp[pr]);
           exit (0);
       }
       pii t=pr;
       t.second = t.first; t.first*=2;
       if (!mp.count(t))
       {
           q.push(t);
           mp[t] = mp[pr]+1;
       }
       t=pr;
       t.first=t.first+t.second;
       if (!mp.count(t))
       {
           q.push(t);
           mp[t] = mp[pr]+1;
       }
   }
   return 0;
}

第三题：

模拟。

#include
using namespace std;
typedef long long ll;
char s[107];
char G[5][10][8] = {
   { "66666" , "....6" , "66666" , "66666" , "6...6" , "66666" , "66666" , "66666" , "66666" , "66666" },
   { "6...6" , "....6" , "....6" , "....6" , "6...6" , "6...." , "6...." , "....6" , "6...6" , "6...6" },
   { "6...6" , "....6" , "66666" , "66666" , "66666" , "66666" , "66666" , "....6" , "66666" , "66666" },
   { "6...6" , "....6" , "6...." , "....6" , "....6" , "....6" , "6...6" , "....6" , "6...6" , "....6" },
   { "66666" , "....6" , "66666" , "66666" , "....6" , "66666" , "66666" , "....6" , "66666" , "66666" }
};
ll cal()
{
   int n = strlen (s);
   ll sum=0, cur=0, prd=1;
   for ( int i=0; i
   {
       if ( isdigit (s[i])) cur=cur*10+s[i]- '0' ;
       else if (s[i] == '-' )
       {
           sum+=prd*cur;
           cur=0;
           prd=-1;
       }
       else if (s[i] == '+' )
       {
           sum+=prd*cur;
           cur=0;
           prd=1;
       }
       else
       {
           prd*=cur;
           cur=0;
       }
   }
   return sum+prd*cur;
}
int main()
{
   int T;
   scanf ( "%d" ,&T); while (T--)
   {
       scanf ( "%s" , s);
       ll ans = cal();
       for ( int i=0; i<5; ++i)
       {
           vector< int > v;
           ll tmp = ans;
           while (tmp) v.push_back(tmp%10),tmp/=10;
           reverse(v.begin(), v.end());
           if (v.empty()) v.push_back(0);
           for ( int j=0; j
           {
               printf ( "%s%s" ,G[i][v[j]], j+1==v.size()? "\n" : ".." );
           }
       }
   }
   return 0;
}

第四题：

只能从均值大的集合往均值小的集合里放。

取数只能取出现次数等于1的数
放数只能放没出现过的数
从小的数开始放可以使均值小的集合均值上升慢，均值大的集合均值上升快，这样最优。

#include
using namespace std;
typedef long long ll;
set< int > sa,sb;
ll suma, sumb;
const long double eps = 1e-14;
inline int cmp( long double a, long double b)
{
   if ( fabs (a-b) <= eps) return 0;
   return a>b?1:-1;
}
inline long double jz(ll k, int m)
{
   return ( long double )k/m;
}
int main()
{
   int n,m;
   scanf ( "%d%d" ,&n,&m);
   for ( int i=0; i
   {
       int t;
       scanf ( "%d" ,&t);
       sa.insert(t);
       suma+=t;
   }
   for ( int i=0;i
   {
       int t;
       scanf ( "%d" ,&t);
       sb.insert(t);
       sumb+=t;
   }
   if (cmp(jz(suma, n), jz(sumb, m))==-1)
   {
       swap(suma, sumb);
       swap(n, m);
       sa.swap(sb);
   }
   int mx =n;
   int ans = 0;
   for (auto k : sa)
   {
       if (cmp(k, jz(suma, n)) >= 0) break ;
//        printf("%d %d\n",n, k);
       if (!sb.count(k)&&cmp(k, jz(sumb, m))>0)
       {
           ++ans;
           sumb+=k;
           suma-=k;
           sb.insert(k);
           --n;++m;
       }
   }
   printf ( "%d\n" , ans);

第五题：BFS

#include
using namespace std;
const int N = 1e5+1000;
typedef pair< int , int > pii;
bool vis[N];
int a[N];
int main()
{
   int n,k,h;
   scanf ( "%d%d%d" ,&n,&k,&h);
   for ( int i=0;i
   {
       int t;
       scanf ( "%d" ,&t);
       a[t]=1;
   }
   queue q;
   q.push({0,0});
   int ans = 0;
   while (!q.empty())
   {
       pii p = q.front(); q.pop();
       if (p.second>k) break ;
       ans = max(ans, p.first);
       for ( int i=1; i<=h; ++i)
       {
           if (a[p.first + i]&&!vis[p.first+2*i])
           {
               vis[p.first+2*i]= true ;
               q.push(make_pair(p.first+2*i, p.second+1));
           }
           if (p.first-2*i>0&&a[p.first-i]&&!vis[p.first-2*i])
           {
               vis[p.first-2*i]= true ;
               q.push(make_pair(p.first-2*i, p.second+1));
           }
       }
   }
   printf ( "%d\n" , ans);
   return 0;
}