Codeforces Round #513 by Barcelona Bootcamp

A. Phone Numbers

签.

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

 #define N 110
 char s[N];
 int cnt[], n;

 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         scanf("%s", s + );
         memset(cnt, , sizeof cnt);
         for (int i = ; i <= n; ++i)
             ++cnt[s[i] - ''];
         int res = ;
         for (int i = ;  i <= cnt[]; ++i)
             res = max(res, min(i, (n - i) / ));
         printf("%d\n", res);
     }
     return ;
 }

B. Maximum Sum of Digits

Solved.

题意：

将一个数拆成$a + b = n$

$要求S(a) + S(b)最大$

$定义S(x) 为x的各位数字之和(以十进制表示)$

思路：

贪心构造一个数字$a有尽量多的9并且小于n即可$

$为什么这样是对的$

我们假设最优解为$a, b$

$令a_i, b_i表示a, b的第i位$

$如果存在某个a_i，不是9， 那么我们将它改成9$

$那么对b_i的影响就是要减去差值$

$如果造成退位，那么答案更优$

$如果没有退位，那么答案不变$

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

 #define ll long long
 ll n;

 int f(ll x)
 {
     int res = ;
     while (x)
     {
         res += x % ;
         x /= ;
     }
     return res;
 }

 int main()
 {
     while (scanf("%lld", &n) != EOF)
     {
         int res = ;
         if (n <= )
         {
             for (int i = ; i <= n; ++i)
                 res = max(res, f(i) + f(n - i));
         }
         else
         {
             ll tmp = n;
             string s = "";
             while (tmp)
             {
                 if (tmp / ) s += '';
                 else if (tmp > ) s += tmp -  + '';
                 tmp /= ;
             }
             ll x = ;
             reverse(s.begin(), s.end());
             for (int i = , len = s.size(); i < len; ++i) x = x *  + s[i] - '';
             res = f(x) + f(n - x);
             ll mid = n / ;
             for (ll i = max(0ll, mid - ); i <= min(n, mid + ); ++i)
                 res = max(res, f(i) + f(n - i));
         }
         printf("%d\n", res);
     }
     return ;
 }

C. Maximum Subrectangle

Solved.

题意：

有一个矩阵$c_{i, j} = a_i \cdot b_j$

$求一个最大子矩阵使得\sum\limits_{i = x_1}^{x_2} \sum\limits_{i = y_1}^{y_2} c_{i, j} <= x$

思路：

我们考虑将求和的式子变换一下

$\sum\limits_{i = x_1}^{x_2} \sum\limits_{i = y_1}^{y_2} c_{i, j} = $

$\sum\limits_{i = x_1}^{x_2} \sum\limits_{i = y_1}^{y_2} a_i \cdot b_j = $

$\sum\limits_{i = x_1}^{x_2} a_i \cdot \sum\limits_{i = y_1}^{y_2} b_j $

那么我们要想使面积最大

只要先预处理出长度为$x的子段和最小的a[] 和 长度为y的字段和最小的b[]$

$再枚举x, y更新答案即可$

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

 #define ll long long
 #define N 2010
 int n, m, a[N], b[N];
 ll x;
 int f[N], g[N];

 int main()
 {
     while (scanf("%d%d", &n, &m) != EOF)
     {
         for (int i = ; i <= n; ++i) scanf("%d", a + i), a[i] += a[i - ];
         for (int i = ; i <= m; ++i) scanf("%d", b + i), b[i] += b[i - ];
         scanf("%lld", &x);
         memset(f, 0x3f, sizeof f);
         memset(g, 0x3f, sizeof g);
         for (int i = ; i <= n; ++i)
             for (int j = i; j <= n; ++j)
                 f[j - i + ] = min(f[j - i + ], a[j] - a[i - ]);
         for (int i = ; i <= m; ++i)
             for (int j = i; j <= m; ++j)
                 g[j - i + ] = min(g[j - i + ], b[j] - b[i - ]);
         int res = ;
         for (int i = ; i <= n; ++i)
             for (int j = ; j <= m; ++j)
                 if (1ll * f[i] * g[j] <= x)
                     res = max(res, i * j);
         printf("%d\n", res);
     }
     return ;
 }

D. Social Circles

Solved.

题意：

有$n个人，要坐在若干个圆桌上，每个人要求左边和右边至少有l_i 和r_i个空凳子$

$空凳子部分可以交叉$

至少需要几张凳子

思路：

我们考虑答案就是$\sum l_i + \sum r_i + n - 交叉部分$

$那么我们就是尽量让交叉部分大即可$

$那么右边空凳子需求多的要和左边空凳子需求多的相邻即可$

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

 #define ll long long
 #define N 100010
 int n;
 int l[N], r[N];

 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         for (int i = ; i <= n; ++i) scanf("%d%d", l + i, r + i);
         sort(l + , l +  + n);
         sort(r + , r +  + n);
         ll res = ;
         for (int i = ; i <= n; ++i) res += max(l[i], r[i]);
         printf("%lld\n", res + n);
     }
     return ;
 }

E. Sergey and Subway

Solved.

题意：

给出一棵树

$现在如果有两个点共同邻接另外一个点$

$那么就给这两个点连一条边$

求新图中所有点对之间的最短距离

思路：

我们考虑新图中两点之间距离和原图之间的关系

$我们考虑下面这样的$

$1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8$

$我们用dis[x] 表示x -> 1的距离$

$在原图中$

$dis_2 = 1, dis_3 = 2, dis_4 = 3, dis_5 = 4, dis_6 = 5, dis_7 = 6, dis_8 = 7$

在新图中

$dis_2 = 1, dis_3 = 1, dis_4 = 2, dis_5 = 2, dis_6 = 3, dis_7 = 3, dis_8 = 4$

$我们发现新图和原图中距离的关系是$

$dis_新 = (dis_原 + 1) / 2$

$用线段树维护原图距离以及奇数个数再树形dp即可$

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

 #define ll long long
 #define N 200010
 int n;
 vector <int> G[N];

 namespace SEG
 {
     struct node
     {
         ll a, b, lazy, cnt;
         void init()
         {
             a = b = lazy = cnt = ;
         }
         void add(ll x)
         {
             a += x * cnt;
             lazy += x;
             if (abs(x) % )
                 b = cnt - b;
         }
         ll f()
         {
             return (a + b) / ;
         }
         node operator + (const node &other) const
         {
             node res; res.init();
             res.a = a + other.a;
             res.b = b + other.b;
             res.cnt = cnt + other.cnt;
             return res;
         }
     }a[N << ];
     void build(int id, int l, int r)
     {
         a[id].init();
         a[id].cnt = r - l + ;
         if (l == r) return;
         int mid = (l + r) >> ;
         build(id << , l, mid);
         build(id <<  | , mid + , r);
     }
     void pushdown(int id)
     {
         if (!a[id].lazy) return;
         a[id << ].add(a[id].lazy);
         a[id <<  | ].add(a[id].lazy);
         a[id].lazy = ;
     }
     void update(int id, int l, int r, int ql, int qr, int val)
     {
         if (l >= ql && r <= qr)
         {
             a[id].add(val);
             return;
         }
         int mid = (l + r) >> ;
         pushdown(id);
         if (ql <= mid) update(id << , l, mid, ql, qr, val);
         if (qr > mid) update(id <<  | , mid + , r, ql, qr, val);
         a[id] = a[id << ] + a[id <<  | ];
     }
 }

 int fa[N], deep[N], in[N], out[N], cnt;
 void DFS(int u)
 {
     in[u] = ++cnt;
     for (auto v : G[u]) if (v != fa[u])
     {
         fa[v] = u;
         deep[v] = deep[u] + ;
         DFS(v);
     }
     out[u] = cnt;
 }

 ll res;
 void DFS2(int u)
 {
     res += SEG::a[].f();
     for (auto v : G[u]) if (v != fa[u])
     {
         SEG::update(, , n, , n, );
         SEG::update(, , n, in[v], out[v], -);
         DFS2(v);
         SEG::update(, , n, , n, -);
         SEG::update(, , n, in[v], out[v], );
     }
 }
 int main()
 {
     while (scanf("%d", &n) != EOF)
     {
         cnt = ; res = ;
         for (int i = ; i <= n; ++i) G[i].clear();
         for (int i = , u, v; i < n; ++i)
         {
             scanf("%d%d", &u, &v);
             G[u].push_back(v);
             G[v].push_back(u);
         }
         deep[] = ;
         DFS();
         SEG::build(, , n);
         for (int i = ; i <= n; ++i) SEG::update(, , n, in[i], in[i], deep[i]);
         DFS2();
         printf("%lld\n", res / );
     }
     return ;
 }