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 = 3 - (n - 22);
            for (int i = need; i; --i) printf(" Eve");
            puts("");
        }
        return 0;
    }

B Christmas Eve Eve

Solved.

    #include <bits/stdc++.h>
    using namespace std;
     
    int main()
    {
        int n;
        while (scanf("%d", &n) != EOF)
        {
            int res = 0, Max = 0;
            for (int i = 1, p; i <= n; ++i)
            {
                scanf("%d", &p);
                res += p;
                Max = max(Max, p);
            }
            printf("%d
", res - Max / 2);
        }
        return 0;
    }

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 = 1; i<= n; ++i) scanf("%d", h + i);
            sort(h + 1, h + 1 + n, [](int a, int b) { return a > b; });
            int res = 1e9;
            for (int i = 1; i + k - 1 <= n; ++i) res = min(res, h[i] - h[i + k - 1]);
            printf("%d
", res);
        }
        return 0;
    }

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
", l, r, cur);
        if (cur < 0 || l > x) return;  
        if (r <= x) 
        {
            res += tot[cur]; 
            return;
        }
        ll mid = (l + r) >> 1;
        if (mid <= x) ++res;
        DFS(l + 1, mid - 1, cur - 1);
        DFS(mid + 1, r - 1, cur - 1);
    }
     
    int main()
    {
        len[0] = 1;
        for (int i = 1; i <= 50; ++i)
            len[i] = 2 * len[i - 1] + 3;
        tot[0] = 1;
        for (int i = 1; i <= 50; ++i)
            tot[i] = 2 * tot[i - 1] + 1;
        while (scanf("%d%lld", &n, &x) != EOF)
        {
            res = 0;
            DFS(1, len[n], n); 
            printf("%lld
", res); 
        }
        return 0;
    }