A Christmas Eve Eve Eve

Solved.

 1     #include <bits/stdc++.h>
 2     using namespace std;
 3      
 4     int main()
 5     {
 6         int n;
 7      
 8         while (scanf("%d", &n) != EOF)
 9         {
10             printf("Christmas"); 
11             int need = 3 - (n - 22
 
);
12             for (int i = need; i; --i) printf(" Eve");
13             puts("");
14         }
15         return 0;
16     }

B Christmas Eve Eve

Solved.

 1     #include <bits/stdc++.h>
 2     using namespace std;
 3      
 4     int main()
 5     {
 
 6         int n;
 7         while (scanf("%d", &n) != EOF)
 8         {
 9             int res = 0, Max = 0;
10             for (int i = 1, p; i <= n; ++i)
11             {
12                 scanf("%d", &p);
13                 res += p;
14                 Max = max(Max, p);
15             }
 
16             printf("%d\n", res - Max / 2);
17         }
18         return 0;
19     }

C Christmas Eve

Solved.

 1     #include <bits/stdc++.h>
 2     using namespace std;
 3      
 4     #define N 100010
 5     int n, k, h[N];
 6      
 7     int main()
 8     {
 9         while (scanf("%d%d", &n, &k) != EOF)
10         {
11             for (int i = 1; i<= n; ++i) scanf("%d", h + i);
12             sort(h + 1, h + 1 + n, [](int a, int b) { return a > b; });
13             int res = 1e9;
14             for (int i = 1; i + k - 1 <= n; ++i) res = min(res, h[i] - h[i + k - 1]);
15             printf("%d\n", res);
16         }
17         return 0;
18     }

D Christmas

Solved.

題意：

遞迴定義了一個漢堡，顯然它是對稱的，求從一端吃掉它長度L，吃掉多少patty

思路：

顯然，漢堡的長度和擁有patty的個數都是可以線性遞推的，先預處理

然後按區間遞迴下去求答案即可。

 1     #include <bits/stdc++.h>
 2     using namespace std;
 3      
 4     #define ll long long
 5     #define N 110
 6     int n; ll x; 
 7     ll len[N], tot[N]; 
 8     ll res;
 9      
10     void DFS(ll l, ll r, int cur)
11     {
12         //printf("%lld %lld %d\n", l, r, cur);
13         if (cur < 0 || l > x) return;  
14         if (r <= x) 
15         {
16             res += tot[cur]; 
17             return;
18         }
19         ll mid = (l + r) >> 1;
20         if (mid <= x) ++res;
21         DFS(l + 1, mid - 1, cur - 1);
22         DFS(mid + 1, r - 1, cur - 1);
23     }
24      
25     int main()
26     {
27         len[0] = 1;
28         for (int i = 1; i <= 50; ++i)
29             len[i] = 2 * len[i - 1] + 3;
30         tot[0] = 1;
31         for (int i = 1; i <= 50; ++i)
32             tot[i] = 2 * tot[i - 1] + 1;
33         while (scanf("%d%lld", &n, &x) != EOF)
34         {
35             res = 0;
36             DFS(1, len[n], n); 
37             printf("%lld\n", res); 
38         }
39         return 0;
40     }