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

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, -, sizeof pos);
     for (int i = ; i >= ; --i)
     {
         for (int j = ; j <= i; ++j) if (i % j == )
         {
             int a = i / j, b = j;
             if (a > b) swap(a, b);
             if (a <=  && b <=  && a <= w[i])
             {
                 pos[i] = i;
                 w[i] = a;
                 h[i] = b;
             }
             if (w[i + ] < w[i])
             {
                 pos[i] = pos[i + ];
                 w[i] = w[i + ];
                 h[i] = h[i + ];
             }
         }
     }
 }

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

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][];

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

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

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-;
 ll a, b, c, x[], y[];
 double xa[], ya[], xb[], yb[];

 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 = ; i < ; ++i) scanf("%lld%lld", x + i, y + i);
         double res = abs(x[] - x[]) + abs(y[] - y[]);
         xa[] = x[]; ya[] = calcy(x[]);
         xa[] = calcx(y[]); ya[] = y[];
         xb[] = x[]; yb[] = calcy(x[]);
         xb[] = calcx(y[]); yb[] = y[];
         for (int i = ; i < ; ++i)
         {
             for (int j = ; j < ; ++j)
             {
                 res = min(res, dis(x[], y[], xa[i], ya[i]) + dis(x[], y[], xb[j], yb[j]) + dis(xa[i], ya[i], xb[j], yb[j]));
             }
         }
         printf("%.15f\n", res);
     }
     return ;
 }

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 = ; i <= ; ++i) for (int j = ; j <= i; ++j)
     {
         if (j ==  || j == i) C[i][j] = ;
         else C[i][j] = C[i - ][j - ] + C[i - ][j];
     }
 }

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

F. Vasya and Maximum Matching

Unsolved.

G. Chattering

Unsolved.