AtCoder Beginner Contest 115 Solution

A Christmas Eve Eve Eve

Solved.

     #include <bits/stdc++.h>
     using namespace std;

     int main()
     {
         int n;

         while (scanf("%d", &n) != EOF)
         {
             printf("Christmas");
             int need =  - (n - );
             for (int i = need; i; --i) printf(" Eve");
             puts("");
         }
         return ;
     }

B Christmas Eve Eve

Solved.

     #include <bits/stdc++.h>
     using namespace std;

     int main()
     {
         int n;
         while (scanf("%d", &n) != EOF)
         {
             int res = , Max = ;
             for (int i = , p; i <= n; ++i)
             {
                 scanf("%d", &p);
                 res += p;
                 Max = max(Max, p);
             }
             printf("%d\n", res - Max / );
         }
         return ;
     }

C Christmas Eve

Solved.

     #include <bits/stdc++.h>
     using namespace std;

     #define N 100010
     int n, k, h[N];

     int main()
     {
         while (scanf("%d%d", &n, &k) != EOF)
         {
             for (int i = ; i<= n; ++i) scanf("%d", h + i);
             sort(h + , h +  + n, [](int a, int b) { return a > b; });
             int res = 1e9;
             for (int i = ; i + k -  <= n; ++i) res = min(res, h[i] - h[i + k - ]);
             printf("%d\n", res);
         }
         return ;
     }

D Christmas

Solved.

题意：

递归定义了一个汉堡，显然它是对称的，求从一端吃掉它长度L，吃掉多少patty

思路：

显然，汉堡的长度和拥有patty的个数都是可以线性递推的，先预处理

然后按区间递归下去求答案即可。

     #include <bits/stdc++.h>
     using namespace std;

     #define ll long long
     #define N 110
     int n; ll x;
     ll len[N], tot[N];
     ll res;

     void DFS(ll l, ll r, int cur)
     {
         //printf("%lld %lld %d\n", l, r, cur);
         if (cur <  || l > x) return;
         if (r <= x)
         {
             res += tot[cur];
             return;
         }
         ll mid = (l + r) >> ;
         if (mid <= x) ++res;
         DFS(l + , mid - , cur - );
         DFS(mid + , r - , cur - );
     }

     int main()
     {
         len[] = ;
         for (int i = ; i <= ; ++i)
             len[i] =  * len[i - ] + ;
         tot[] = ;
         for (int i = ; i <= ; ++i)
             tot[i] =  * tot[i - ] + ;
         while (scanf("%d%lld", &n, &x) != EOF)
         {
             res = ;
             DFS(, len[n], n);
             printf("%lld\n", res);
         }
         return ;
     }