Avito Cool Challenge 2018 Solution

A. Definite Game

签.

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

 int main()
 {
     int a;
     while (scanf("%d", &a) != EOF)
     {
         [](int x)
         {
             for (int i = x - ; i >= ; --i) if (x % i)
             {
                 printf("%d\n", x - i);
                 return;
             }
             printf("%d\n", x);
         }(a);
     }
     return ;
 }

B. Farewell Party

签。

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

 #define N 100010
 int n, a[N], b[N], cnt[N];
 vector <int> v[N];
 int ans[N];

 bool ok()
 {
     for (int i = ; i <= n; ++i) if (!v[i].empty() && v[i].size() != i)
         return false;
     return true;
 }
 void solve()
 {
     int cnt = ;
     for (int i = ; i <= n; ++i) v[i].clear();
     for (int i = ; i <= n; ++i)
     {
         int id = n - a[i];
         if (v[id].size() == id)
         {
             ++cnt;
             for (auto it : v[id]) ans[it] = cnt;
             v[id].clear();
         }
         v[id].push_back(i);
     }
     if (!ok())
     {
         puts("Impossible");
         return;
     }
     else
     {
         for (int i = ; i <= n; ++i) if (!v[i].empty())
         {
             ++cnt;
             for (auto it : v[i])
                 ans[it] = cnt;
         }
         puts("Possible");
         for (int i = ; i <= n; ++i) printf("%d%c", ans[i], " \n"[i == n]);
     }
 }

 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         for (int i = ; i <= n; ++i) scanf("%d", a + i);
         solve();
     }
     return ;
 }

C. Colorful Bricks

Upsolved.

题意：

有$一行n个方格，m种颜色，恰好有k个方块与它左边方块的颜色不同$

求方案数

思路：

$dp[i][j] 表示到第i个方格，有j个方块与左边方块颜色不同的方案数$

转移有

$dp[i + 1][j] += dp[i][j]$

$dp[i + 1][j + 1] += dp[i][j] * (m - 1)$

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

 #define ll long long
 #define N 2010
 const ll MOD = (ll);
 int n, m, k;
 ll f[N][N];

 int main()
 {
     while (scanf("%d%d%d", &n, &m, &k) != EOF)
     {
         memset(f, , sizeof f);
         f[][] = m;
         for (int i = ; i <= n; ++i)
         {
             for (int j = ; j <= min(i - , k); ++j)
             {
                 f[i][j] = (f[i][j] + f[i - ][j]) % MOD;
                 f[i][j + ] = (f[i][j + ] + f[i - ][j] * (m - ) % MOD) % MOD;
             }
         }
         printf("%lld\n", f[n][k]);
     }
     return ;
 }

D. Maximum Distance

Upsolved.

题意：

有一张图

定义一条路径的长度的路径上边权最大值

定义两点距离为两点之间最短路径

有一些特殊点，要对所有特殊点求离它最远的特殊点

思路：

我们考虑一条边所连接的两个连通块里，如果这两个连通块里都有特殊点

那么这条边的权值就可以用于更新特殊点的答案

其实就是一个求最小生成树的过程，先按边权排序

要注意的是不是求整个图的最小生成树，而是所有特殊点的最小生成树

其实是最小瓶颈生成树.

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

 #define N 100010
 int n, m, k;
 struct node
 {
     int u, v, w;
     void scan() { scanf("%d%d%d", &u, &v, &w); }
     bool operator < (const node &other) const { return w < other.w; }
 }edge[N];

 int pre[N], cnt[N];
 int find(int x) { return pre[x] ==  ? x : pre[x] = find(pre[x]); }
 int Kruskal()
 {
     sort(edge + , edge +  + m);
     for (int i = ; i <= m; ++i)
     {
         int u = edge[i].u, v = edge[i].v, w = edge[i].w;
         int fu = find(u), fv = find(v);
         if (fu == fv) continue;
         cnt[fv] += cnt[fu];
         pre[fu] = fv;
         if (cnt[fv] == k) return w;
     }
 }

 int main()
 {
     while (scanf("%d%d%d", &n, &m, &k) != EOF)
     {
         memset(pre, , sizeof pre);
         memset(cnt, , sizeof cnt);
         for (int i = , x; i <= k; ++i) scanf("%d", &x), cnt[x] = ;
         for (int i = ; i <= m; ++i) edge[i].scan();
         int res = Kruskal();
         for (int i = ; i <= k; ++i) printf("%d%c", res, " \n"[i == k]);
     }
     return ;
 }

E. Missing Numbers

Upsolved.

题意：

有$n个数，给出a_2, a_4 \cdots a_n$

$n是偶数，提供a_1, a_3 \cdots a_{n - 1}$

使得

$所有前缀和都是平方数$

思路：

考虑$n = 2的时候$

$我们令b^2 = a_1 + a_2$

$a^2 = a_1$

那么有

$b^2 - a^2 = (b + a) \cdot (b - a) = a_2$

$我们将a_2 拆成 p \cdot q 的形式$

$令p >= q$

$那么有  b + a = p, b - a = q$

$解方程有 b = \frac{p + q}{2}$

$发现一限制条件 p, q的奇偶性要相同$

$那么a_1 = (\frac{p + q}{2})^2 - a_2$

那么如果有多解 我们取$a_1最小的$

$因为当n >= 2的情况也可以同样这样推下去，会发现前面的项越小越好$

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

 #define ll long long
 #define N 100010
 int n;
 ll x[N];
 vector <int> vec[N << ];

 void init()
 {
     for (int i = ; i <= ; ++i)
         for (int j = i * i; j <= ; j += i)
             vec[j].push_back(i);
 }

 int main()
 {
     init();
     while (scanf("%d", &n) != EOF)
     {
         for (int i = ; i <= n; i += ) scanf("%lld", x + i);
         bool flag = true;
         ll base = ;
         for (int i = ; i <= n; i += )
         {
             base += x[i + ];
             int limit = sqrt(x[i + ]);
             ll res = (ll)1e18;
             ll p, q;
             for (auto j : vec[x[i + ]])
                 if ((j % ) == ((x[i + ] / j) % ))
                 {
                     p = j, q = (x[i + ] / j);
                     ll tot = (p + q) / ;
                     tot *= tot;
                     if (tot > base)
                         res = min(res, tot - base);
                 }
             if (res <= (ll)1e13)
             {
                 x[i] = res;
                 base += res;
             }
             else
             {
                 flag = false;
                 break;
             }
         }
         if (!flag) puts("No");
         else
         {
             puts("Yes");
             for (int i = ; i <= n; ++i)
                 printf("%lld%c", x[i], " \n"[i == n]);
         }
     }
     return ;
 }