Codeforces Round #522 (Div. 2, based on Technocup 2019 Elimination Round 3) Solution

A. Kitchen Utensils

Water.

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

#define N 110
int n, k, cnt[N];

int main()
{
    while (scanf("%d%d", &n, &k) != EOF)
    {
        memset(cnt, 0, sizeof cnt);
        for (int i = 1, x; i <= n; ++i) 
        {
            scanf("%d", &x);
            ++cnt[x];
        }
        int Max = 0;
        for (int i = 1; i <= 100; ++i) Max = max(Max, cnt[i] % k == 0 ? cnt[i] / k : cnt[i] / k + 1);
        int res = 0;
        for (int i = 1; i <= 100; ++i) if (cnt[i])
        {
            res += Max * k - cnt[i];
        }
        printf("%d\n", res);
    }
    return 0;
}

 

B. Personalized Cup

Water.

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

#define INF 0x3f3f3f3f
#define N 110
int pos[N], w[N], h[N], n;
string s;

void init()
{
    memset(w, 0x3f, sizeof w);
    memset(h, 0x3f, sizeof h);
    memset(pos, -1, sizeof pos); 
    for (int i = 100; i >= 1; --i)
    {
        for (int j = 1; j <= i; ++j) if (i % j == 0)
        {
            int a = i / j, b = j;
            if (a > b) swap(a, b);
            if (a <= 5 && b <= 20 && a <= w[i]) 
            {
                pos[i] = i;
                w[i] = a;
                h[i] = b; 
            }
            if (w[i + 1] < w[i])
            {
                pos[i] = pos[i + 1];
                w[i] = w[i + 1];
                h[i] = h[i + 1];
            }
        }
    }
}

int main()
{
    init();
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    while (cin >> s)
    {
        n = s.size();
        cout << w[n] << " " << h[n] << "\n"; 
        int need = pos[n] - n;
        int must = need / w[n];
        int remind = need % w[n];
        for (int i = 1, pos = 0; i <= w[n]; ++i)
        {
            int j = 1;
            for (; j <= must; ++j) cout << "*";
            if (remind) cout << "*", ++j, --remind;
            for (; j <= h[n]; ++j) cout << s[pos++];
            cout << "\n";
        }
    }
    return 0;
}

 

C. Playing Piano

Upsolved.

思路：每一位枚举，记忆化搜索。$dp[i][j] 表示第i位放j是否可以$

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

#define N 100010
int n, a[N], res[N], dp[N][10];

int DFS(int i, int num)
{
    if (i > n) return 1;
    int &x = dp[i][num];
    if (x != -1) return x;
    if (a[i] > a[i - 1])
    {
        for (int j = num + 1; j <= 5; ++j) 
        {
            x = DFS(i + 1, j);
            if (x) return res[i] = j, x = 1;
        }
    }    
    else if (a[i] < a[i - 1])
    {
        for (int j = 1; j < num; ++j)
        {
            x = DFS(i + 1, j);
            if (x) return res[i] = j, x = 1;
        }
    }
    else
    {
        for (int j = 1; j <= 5; ++j) if (j != num)
        {
            x = DFS(i + 1, j);
            if (x)  return res[i] = j, x = 1;
        }
    }
    return x = 0;
}

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        memset(dp, -1, sizeof dp);
        a[0] = 0;
        for (int i = 1; i <= n; ++i) scanf("%d", a + i);
        if (DFS(1, 0)) for (int i = 1; i <= n; ++i) printf("%d%c", res[i], " \n"[i == n]);
        else puts("-1");    
    }
    return 0;
}

 

D. Barcelonian Distance

Solved.

题意：

在一个二维平面中，有一条直线，有两点个点，只能走这条直线上或者方格线上，求两点最短路径

思路：

考虑每个点到直线上通过方格线走只有两种方式，即横着走，纵着走

那么通过直线的方式即有四种

还有一种直接是曼哈顿距离，取$Min$即可

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

#define ll long long
const double eps = 1e-8;
ll a, b, c, x[2], y[2];
double xa[2], ya[2], xb[2], yb[2];

double dis(double x1, double y1, double x2, double y2)
{
    return sqrt((x1 - x2) * (x1 - x2) * 1.0 + (y1 - y2) * (y1 - y2) * 1.0);  
}

double calcx(ll y) 
{
    return -(b * y + c) * 1.0  / a; 
}

double calcy(ll x)
{
    return -(a * x + c) * 1.0  / b; 
}

int main()
{
    while (scanf("%lld%lld%lld", &a, &b, &c) != EOF)
    {
        for (int i = 0; i < 2; ++i) scanf("%lld%lld", x + i, y + i);
        double res = abs(x[0] - x[1]) + abs(y[0] - y[1]); 
        xa[0] = x[0]; ya[0] = calcy(x[0]);
        xa[1] = calcx(y[0]); ya[1] = y[0];
        xb[0] = x[1]; yb[0] = calcy(x[1]);
        xb[1] = calcx(y[1]); yb[1] = y[1];
        for (int i = 0; i < 2; ++i)
        {
            for (int j = 0; j < 2; ++j)
            {
                res = min(res, dis(x[0], y[0], xa[i], ya[i]) + dis(x[1], y[1], xb[j], yb[j]) + dis(xa[i], ya[i], xb[j], yb[j]));
            }
        }
        printf("%.15f\n", res);
    }
    return 0;
}

 

 

E. The Unbearable Lightness of Weights

Upsolved.

题意：

有n个砝码，知道所有砝码的重量，但是不知道每个砝码和重量的对应关系，可以有一次询问，问用k个砝码组成总重量为m的方案，求最后你最多知道多少个砝码的具体重量

思路：

如果不同重量的砝码种类不超过2，直接输出n

01背包求出拼成重量为w, 物品个数为k的方案数，然后对于每种重量的砝码枚举个数，当且仅当当前个数和数量只有唯一一种方式组成时，那么这个数量就是合法的，可以用来更新答案

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

#define N 110
int n, sum, tot, a[N], dp[N * N][N], cnt[N], C[N][N];

void init()
{
    for (int i = 0; i <= 100; ++i) for (int j = 0; j <= i; ++j)
    {
        if (j == 0 || j == i) C[i][j] = 1;
        else C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
}

void Run()
{
    init();
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; ++i)
        {
            scanf("%d", a + i), sum += a[i], ++cnt[a[i]];
            if (cnt[a[i]] == 1) ++tot;
        }
        if (tot <= 2)
        {
            printf("%d\n", n);
            continue;
        }
        dp[0][0] = 1;
        for (int i = 1; i <= n; ++i) 
        {
            for (int j = sum; j >= a[i]; --j) for (int k = 1; k <= n; ++k)
                dp[j][k] += dp[j - a[i]][k - 1];
        }
        int res = 0; 
        for (int i = 1; i <= 100; ++i) if (cnt[i])
        {
            for (int j = 1; j <= cnt[i]; ++j)
            {
                if (dp[i * j][j] == C[cnt[i]][j])
                    res = max(res, j);
            }
        }
        printf("%d\n", res);
    }
}

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif 

    Run();
    return 0;
}

 

F. Vasya and Maximum Matching

Unsolved.

 

G. Chattering

Unsolved.