标签：can space base 还原 inf inline fibonacci enc for
目录
| Solved | A | B | C | D | E | F | G | H | I | J | K | L | M | N |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 9/14 | O | O | - | O | - | O | O | O | O | - | O | - | - | O |
题意：
给出一个\(n \cdot n\)的矩形，这个矩形\(a_{i, j}\)的初始值为\(0\)，它每次能够选择一行或者一列加上\(1\)，现在遮住某个位置的数，让你还原这个数。
思路：
考虑倒退操作，不考虑遮住的那个数，然后枚举每行，每列，每次选择行最小，列最小将整行整列减去即可还原出那个数。
代码：
view code
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
int n, a[N][N];
int main() {
while (scanf("%d", &n) != EOF) {
int x = -1, y = -1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
scanf("%d", &a[i][j]);
if (a[i][j] == -1) {
x = i, y = j;
a[i][j] = 0;
}
}
}
for (int i = 1; i <= n; ++i) {
int Min = 1e9;
for (int j = 1; j <= n; ++j) {
if (x == i && y == j) continue;
Min = min(Min, a[i][j]);
}
for (int j = 1; j <= n; ++j) {
a[i][j] -= Min;
}
}
for (int j = 1; j <= n; ++j) {
int Min = 1e9;
for (int i = 1; i <= n; ++i) {
if (x == i && y == j) continue;
Min = min(Min, a[i][j]);
}
for (int i = 1; i <= n; ++i) {
a[i][j] -= Min;
}
}
printf("%d\n", -a[x][y]);
}
return 0;
}题意：
定义一个序列\((a_1, a_2, \cdots, a_n)\)是一个\((n, m, d)-good\)当且仅当\(1 \leq a_i \leq m(1 \leq i \leq n)\)并且\(gcd(a_1, a_2, \cdots, a_n) = d\)
令\(f(a, k) = (a_1a_2\cdots a_n)^k\)，现在给出\(n, m, d, k\)，让你求所有合法的\((n, m, d)-good\)的序列\(a\)的\(f(a, k)\)
思路：
题目要求的东西等价于：
\[
\begin{eqnarray*}
f(d) = \sum\limits_{a_1 = 1}^m \sum\limits_{a_2 = 1}^m \cdots \sum\limits_{a_n = 1}^m [gcd(a_1, a_2, \cdots, a_n) = d](a_1a_2\cdots a_n)^k
\end{eqnarray*}
\]
那么我们令：
\[
\begin{eqnarray*}
g(d) = \sum\limits_{a_1 = 1}^m \sum\limits_{a_2 = 1}^m \cdots \sum\limits_{a_n = 1}^m [d | gcd(a_1, a_2, \cdots, a_n)](a_1a_2\cdots a_n)^k
\end{eqnarray*}
\]
显然有：
\[
\begin{eqnarray*}
g(d) = (\sum\limits_{d\;|\;i} i^k)^n
\end{eqnarray*}
\]
莫比乌斯反演有：
\[
\begin{eqnarray*}
f(d) &=& \sum\limits_{d\;|\;i} \mu(\frac{i}{d})g(i) \&=& \sum\limits_{d\;|\;i} \mu(\frac{i}{d}) (\sum\limits_{i\;|\;j} j^k)^n
\end{eqnarray*}
\]
所以：
\[
f(d) = \sum\limits_{i = 1}^{\left\lfloor m/d \right\rfloor} \mu(i) (\sum\limits_{id\;|\;j} j^k)^n
\]
代码：
view code
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e5 + 10, mod = 59964251;
int pri[N], check[N], mu[N], n, m, d, K, phi;
char s[N];
void sieve() {
memset(check, 0, sizeof check);
*pri = 0;
mu[1] = 1;
for (int i = 2; i < N; ++i) {
if (check[i] == 0) {
pri[++*pri] = i;
mu[i] = -1;
}
for (int j = 1; j <= *pri; ++j) {
if (i * pri[j] >= N) break;
check[i * pri[j]] = 1;
if (i % pri[j] == 0) {
mu[i * pri[j]] = 0;
break;
} else {
mu[i * pri[j]] = -mu[i];
}
}
}
}
int eular(int n) {
int ans = n;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
ans -= ans / i;
while (n % i == 0)
n /= i;
}
}
if (n > 1) ans -= ans / n;
return ans;
}
ll gcd(ll a, ll b) {
return b ? gcd(b, a % b) : a;
}
ll qmod(ll base, ll n) {
ll res = 1;
while (n) {
if (n & 1) res = res * base % mod;
base = base * base % mod;
n >>= 1;
}
return res;
}
int getMod(int mod) {
int res = 0;
for (int i = 1; s[i]; ++i) {
res = (res * 10 + s[i] - '0') % mod;
}
return res;
}
int main() {
phi = eular(mod);
sieve();
// cout << phi << endl;
int _T; scanf("%d", &_T);
while (_T--) {
scanf("%s%d%d%d", s + 1, &m, &d, &K);
int len = strlen(s + 1);
if (len <= 9) {
n = 0;
for (int i = 1; s[i]; ++i) {
n = n * 10 + s[i] - '0';
}
} else {
n = getMod(phi);
if (getMod(643) == 0 || getMod(93257) == 0) {
n += phi;
}
}
ll res = 0;
for (int i = 1; i <= m / d; ++i) {
int base = 0;
for (int j = i * d; j <= m; j += i * d) {
base += qmod(j, K);
base %= mod;
}
res += 1ll * mu[i] * qmod(base, n) % mod;
res = (res + mod) % mod;
}
printf("%lld\n", res);
}
return 0;
}题意：
定义一个multiset的权值为里面任意两个数的异或和的平方的和。
现在给出一棵有根树(\(1\)为根)，每个点有点权，定义\(p(x, k)\)为\(x\)子树中距离\(x\)不超过\(k\)的所有点的点权构成的multiset的权值，现在要对每个\(i \in [1, n]\)求\(p(i, k)\)
裸的线段树。
代码：
view code
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, q;
struct SEG {
struct node {
int Max, lazy;
node() { Max = lazy = 0; }
void up(int x) {
Max += x;
lazy += x;
}
node operator + (const node &other) const {
node res = node();
res.Max = max(Max, other.Max);
return res;
}
}t[N << 2];
void build(int id, int l, int r) {
t[id] = node();
if (l == r) return;
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
}
void down(int id) {
int &lazy = t[id].lazy;
if (lazy) {
t[id << 1].up(lazy);
t[id << 1 | 1].up(lazy);
lazy = 0;
}
}
void update(int id, int l, int r, int ql, int qr, int v) {
if (l >= ql && r <= qr) {
t[id].up(v);
return;
}
int mid = (l + r) >> 1;
down(id);
if (ql <= mid) update(id << 1, l, mid, ql, qr, v);
if (qr > mid) update(id << 1 | 1, mid + 1, r, ql, qr, v);
t[id] = t[id << 1] + t[id << 1 | 1];
}
int query(int id, int l, int r, int ql, int qr) {
if (l >= ql && r <= qr) return t[id].Max;
int mid = (l + r) >> 1;
down(id);
int res = 0;
if (ql <= mid) res = max(res, query(id << 1, l, mid, ql, qr));
if (qr > mid) res = max(res, query(id << 1 | 1, mid + 1, r, ql, qr));
return res;
}
}seg[4];
int main() {
int id[] = {0, 0, 0, 1, 0, 2, 0, 3, 0};
vector <vector<int>> vec;
vec.resize(15);
for (int i = 2; i <= 10; ++i) {
int x = i;
vec[i].clear();
for (int j = 2; j <= x; ++j) {
while (x % j == 0) {
vec[i].push_back(j);
x /= j;
}
}
// cout << i << endl;
// for (auto &it : vec[i])
// cout << it << " ";
// cout << endl;
}
while (scanf("%d%d", &n, &q) != EOF) {
for (int i = 0; i < 4; ++i) seg[i].build(1, 1, n);
char op[20]; int l, r, x;
while (q--) {
scanf("%s%d%d", op, &l, &r);
if (op[1] == 'U') {
scanf("%d", &x);
for (auto &it : vec[x]) {
seg[id[it]].update(1, 1, n, l, r, 1);
}
} else {
int res = 0;
for (int i = 0; i < 4; ++i) {
res = max(res, seg[i].query(1, 1, n, l, r));
}
printf("ANSWER %d\n", res);
}
}
}
return 0;
}题意：
给出一张图，有\(x\)条无向边，有\(y\)条有向边，保证无向边都是正权值，有向边可能有负权值，并且保证如果一条有向边\(a_i \rightarrow b_i\)，那么在该图中，\(b_i\)不可能到达\(a_i\)
现在询问从\(s\)出发到任意一点的最短路。
思路：
我们考虑如果只考虑有向边，那么是一个\(DAG\)，那么把无向边连成的每个联通块看成一个新点，并且有有向边将他们连接起来，他们也是一个\(DAG\)。
并且无向图的连通块里面没有负权边，可以跑dijkstra，然后根据拓扑序dp一下即可。
代码：
view code
#include <bits/stdc++.h>
using namespace std;
using pII = pair<int, int>;
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts>
void err(const T& arg, const Ts&... args) { cout << arg << ' '; err(args...); }
#define fi first
#define se second
const int N = 5e4 + 10, INF = 0x3f3f3f3f;
int n, mx, my, s, id[N], d[N];
//0 two-way 1 one-way
vector <vector<pII>> G[3];
vector <vector<int>> po;
struct DSU {
int fa[N];
void init() { memset(fa, 0, sizeof fa); }
int find(int x) {
return fa[x] == 0 ? x : fa[x] = find(fa[x]);
}
void merge(int u, int v) {
int fu = find(u), fv = find(v);
if (fu != fv) {
fa[fu] = fv;
}
}
}dsu;
struct node {
int u, w;
node() {}
node(int u, int w) : u(u), w(w) {}
bool operator < (const node &other) const {
return w > other.w;
}
};
int dis[N], used[N];
void Dijkstra(int S) {
priority_queue <node> pq;
pq.push(node(S, dis[S]));
while (!pq.empty()) {
int u = pq.top().u; pq.pop();
for (auto &it : G[0][u]) {
int v = it.fi, w = it.se;
if (dis[v] > dis[u] + w) {
dis[v] = dis[u] + w;
pq.push(node(v, dis[v]));
}
}
}
}
void Topo() {
queue <int> que;
for (int i = 1; i <= *id; ++i) {
if (d[i] == 0) {
que.push(i);
}
}
while (!que.empty()) {
int u = que.front(); que.pop();
sort(po[u].begin(), po[u].end(), [&](int a, int b) { return dis[a] < dis[b]; });
for (auto &it : po[u]) {
Dijkstra(it);
for (auto &it2 : G[1][it]) {
int v = it2.fi, w = it2.se;
if (dis[it] < INF) {
dis[v] = min(dis[v], dis[it] + w);
}
}
}
for (auto &it : G[2][u]) {
int v = it.fi;
if (--d[v] == 0) {
que.push(v);
}
}
}
}
int main() {
while (scanf("%d%d%d%d", &n, &mx, &my, &s) != EOF) {
for (int i = 1; i <= n; ++i) {
dis[i] = INF;
}
dis[s] = 0;
*id = 0;
G[0].clear(); G[1].clear();
G[0].resize(n + 1); G[1].resize(n + 1);
for (int i = 1, u, v, w; i <= mx; ++i) {
scanf("%d%d%d", &u, &v, &w);
G[0][u].push_back(pII(v, w));
G[0][v].push_back(pII(u, w));
}
for (int i = 1, u, v, w; i <= my; ++i) {
scanf("%d%d%d", &u, &v, &w);
G[1][u].push_back(pII(v, w));
}
dsu.init();
for (int u = 1; u <= n; ++u) {
for (auto &it : G[0][u]) {
int v = it.fi;
dsu.merge(u, v);
}
}
for (int u = 1; u <= n; ++u) {
if (dsu.fa[u] == 0)
id[u] = ++*id;
}
for (int u = 1; u <= n; ++u) {
if (dsu.fa[u]) {
id[u] = id[dsu.find(u)];
}
}
// for (int i = 1; i <= n; ++i)
// dbg(i, id[i]);
po.clear(); po.resize(*id + 10);
G[2].clear(); G[2].resize(*id + 10);
memset(d, 0, sizeof d);
for (int u = 1; u <= n; ++u) {
po[id[u]].push_back(u);
for (auto &it : G[1][u]) {
int v = it.fi;
if (id[u] != id[v]) {
G[2][id[u]].push_back(pII(id[v], v));
++d[id[v]];
}
}
}
Topo();
for (int i = 1; i <= n; ++i) {
if (dis[i] >= INF) puts("NO PATH");
else printf("%d\n", dis[i]);
}
}
return 0;
}题意：
进制转换
代码：
view code
def main():
l = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L',
'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h',
'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
x, y, z = input().split()
x = int(x)
y = int(y)
num = 0
for c in z:
if ord('0') <= ord(c) <= ord('9'):
num = num * x + ord(c) - ord('0')
elif ord('A') <= ord(c) <= ord('Z'):
num = num * x + ord(c) - ord('A') + 10
else:
num = num * x + ord(c) - ord('a') + 36
if num == 0:
print("0")
return
res = ""
while num > 0:
tmp = num % y
res = res + l[tmp]
num = num // y
# print(res)
res = res[::-1]
print(res)
main()题意：
给出两个\(n \cdot m\)的矩形，并且里面的数是\([1, nm]\)的排列。
求两个矩形的最大公共子矩形，这里的大定义为面积。
思路：
考虑里面的数是一个排列，可以\(O(n^2)\)处理出每个数向上拓展多少，向左右拓展多少，然后考虑每一行，肯定是某段连续的长度然后乘上这段最小的向上拓展的数量。
向上拓展的这部分当成一个维，位置当成一维，丢进笛卡尔树里面跑一跑能跑出每个最小值管辖的范围，然后和左右扩展的范围求交，就是每个向上拓展数量在横向最远能扩展的范围。
代码：
view code
#include <bits/stdc++.h>
using namespace std;
using pII = pair<int, int>;
#define dbg(x...) do { cout << "\033[32;1m" << #x << " -> "; err(x); } while (0)
void err() { cout << "\033[39;0m" << endl; }
template <class T, class... Ts>
void err(const T& arg, const Ts&... args) { cout << arg << ' '; err(args...); }
#define fi first
#define se second
const int N = 1e3 + 10, INF = 0x3f3f3f3f;
int n, m, a[N][N], b[N][N], l[N][N], r[N][N], num[N][N];
pII c[N], id[N * N];
struct CT {
struct node {
int id, val, fa;
int son[2];
node() {}
node (int id, int val, int fa) : id(id), val(val), fa(fa) {
son[0] = son[1] = 0;
}
bool operator < (const node &other) const {
return val < other.val;
}
}t[N];
int root;
void init() {
t[0] = node(0, -INF, 0);
}
void build(int n, int *a) {
for (int i = 1; i <= n; ++i) {
t[i] = node(i, a[i], 0);
}
for (int i = 1; i <= n; ++i) {
int k = i - 1;
while (t[i] < t[k]) {
k = t[k].fa;
}
t[i].son[0] = t[k].son[1];
t[k].son[1] = i;
t[i].fa = k;
t[t[i].son[0]].fa = i;
}
root = t[0].son[1];
}
int dfs(int u) {
if (!u) return 0;
c[u].fi = dfs(t[u].son[0]);
c[u].se = dfs(t[u].son[1]);
return c[u].fi + c[u].se + 1;
}
}ct;
int main() {
while (scanf("%d%d", &n, &m) != EOF) {
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
scanf("%d", &a[i][j]);
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
scanf("%d", &b[i][j]);
id[b[i][j]] = pII(i, j);
}
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
if (j == 1) {
l[i][j] = j;
} else {
int pre = a[i][j - 1], now = a[i][j];
if (id[pre].fi == id[now].fi && id[pre].se == id[now].se - 1) {
l[i][j] = l[i][j - 1];
} else {
l[i][j] = j;
}
}
}
for (int j = m; j >= 1; --j) {
if (j == m) {
r[i][j] = j;
} else {
int nx = a[i][j + 1], now = a[i][j];
if (id[nx].fi == id[now].fi && id[nx].se == id[now].se + 1) {
r[i][j] = r[i][j + 1];
} else {
r[i][j] = j;
}
}
}
}
for (int j = 1; j <= m; ++j) {
for (int i = 1; i <= n; ++i) {
if (i == 1) {
num[i][j] = 1;
} else {
int pre = a[i - 1][j], now = a[i][j];
if (id[pre].fi == id[now].fi - 1 && id[pre].se == id[now].se) {
num[i][j] = num[i - 1][j] + 1;
} else {
num[i][j] = 1;
}
}
}
}
int res = 0;
for (int i = 1; i <= n; ++i) {
// for (int j = 1; j <= m; ++j) {
// dbg(i, j, l[i][j], r[i][j], num[i][j]);
// }
ct.init();
ct.build(n, num[i]);
ct.dfs(ct.root);
for (int j = 1; j <= m; ++j) {
int tl = max(l[i][j], j - c[j].fi);
int tr = min(r[i][j], j + c[j].se);
res = max(res, (tr - tl + 1) * num[i][j]);
}
}
printf("%d\n", res);
}
return 0;
}纯输出题
2019 ICPC Asia Yinchuan Regional
标签：can space base 还原 inf inline fibonacci enc for
原文地址：https://www.cnblogs.com/Dup4/p/11963769.html
