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 - 1; i >= 1; --i) if (x % i)
            {
                printf("%d
", x - i);
                return;
            }
            printf("%d
", x);
        }(a);
    }
    return 0;
}

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 = 1; i <= n; ++i) if (!v[i].empty() && v[i].size() != i)
        return false;
    return true;
}
void solve()    
{    
    int cnt = 0;
    for (int i = 1; i <= n; ++i) v[i].clear();
    for (int i = 1; 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 = 1; i <= n; ++i) if (!v[i].empty()) 
        {
            ++cnt;
            for (auto it : v[i])
                ans[it] = cnt;
        }
        puts("Possible");
        for (int i = 1; i <= n; ++i) printf("%d%c", ans[i], " 
"[i == n]);
    }
}

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; ++i) scanf("%d", a + i);
        solve();
    }
    return 0;
}

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)998244353;
int n, m, k;
ll f[N][N];

int main()
{
    while (scanf("%d%d%d", &n, &m, &k) != EOF)
    {
        memset(f, 0, sizeof f);
        f[1][0] = m;
        for (int i = 2; i <= n; ++i)
        { 
            for (int j = 0; j <= min(i - 2, k); ++j)
            {
                f[i][j] = (f[i][j] + f[i - 1][j]) % MOD;
                f[i][j + 1] = (f[i][j + 1] + f[i - 1][j] * (m - 1) % MOD) % MOD;          
            }
        }
        printf("%lld
", f[n][k]);
    }
    return 0;
}

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] == 0 ? x : pre[x] = find(pre[x]); }
int Kruskal()
{
    sort(edge + 1, edge + 1 + m);
    for (int i = 1; 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, 0, sizeof pre);
        memset(cnt, 0, sizeof cnt); 
        for (int i = 1, x; i <= k; ++i) scanf("%d", &x), cnt[x] = 1;
        for (int i = 1; i <= m; ++i) edge[i].scan();
        int res = Kruskal();
        for (int i = 1; i <= k; ++i) printf("%d%c", res, " 
"[i == k]);
    }
    return 0;
}

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 << 1];

void init()
{
    for (int i = 1; i <= 450; ++i)
        for (int j = i * i; j <= 200000; j += i)
            vec[j].push_back(i);
}
    
int main()
{
    init();
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 2; i <= n; i += 2) scanf("%lld", x + i);
        bool flag = true; 
        ll base = 0;
        for (int i = 1; i <= n; i += 2)
        {
            base += x[i + 1];
            int limit = sqrt(x[i + 1]);
            ll res = (ll)1e18; 
            ll p, q;
            for (auto j : vec[x[i + 1]])                
                if ((j % 2) == ((x[i + 1] / j) % 2))
                {    
                    p = j, q = (x[i + 1] / j);
                    ll tot = (p + q) / 2;
                    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 = 1; i <= n; ++i)
                printf("%lld%c", x[i], " 
"[i == n]);
        }
    }
    return 0;
}