2019杭电多校第二场
2019杭电多校第二场
太菜了，被学弟暴打
1002. Beauty of Unimodal Sequence
upsolved
要求输出字典序最小和最大的最长单峰子序列
对于每一个位置，维护以这个位置结尾的前缀/后缀最长上升/单峰子序列长度，然后贪心输出（如果只要求长度正反求LIS就好了)
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
int n, a[N];
struct Seg_Tree {
int mx[N << 2];
void init(int rt = 1, int l = 1, int r = n + 2) {
mx[rt] = 0;
if(l == r) return;
int mid = l + r >> 1;
init(rt << 1, l, mid);
init(rt << 1 | 1, mid + 1, r);
}
void pushup(int rt) {
mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
}
void update(int rt, int l, int r, int pos, int val) {
if(l == r) {
mx[rt] = max(mx[rt], val);
return;
}
int mid = l + r >> 1;
if(pos <= mid)
update(rt << 1, l, mid, pos, val);
else
update(rt << 1 | 1, mid + 1, r, pos, val);
pushup(rt);
}
int query(int rt, int l, int r, int L, int R) {
if(L <= l && r <= R)
return mx[rt];
int mid = l + r >> 1, ans = 0;
if(L <= mid)
ans = max(ans, query(rt << 1, l, mid, L, R));
if(R > mid)
ans = max(ans, query(rt << 1 | 1, mid + 1, r, L, R));
return ans;
}
}zero, one;
int dp[2][N][2], len;
vector<int> ans[2], all, up[N], down[N];
void init() {
len = 0;
zero.init();
one.init();
all.clear();
ans[0].clear();
ans[1].clear();
for(int i = 0; i <= n + 1; ++i)
up[i].clear(), down[i].clear();
}
int main() {
while(~scanf("%d", &n)) {
init();
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
all.push_back(a[i]);
}
all.push_back(0);
all.push_back(0x3f3f3f3f);
sort(all.begin(), all.end());
all.erase(unique(all.begin(), all.end()), all.end());
for(int i = 1; i <= n; ++i) {
a[i] = lower_bound(all.begin(), all.end(), a[i]) - all.begin() + 1;
}
for(int i = 1; i <= n; ++i) {
dp[0][i][0] = zero.query(1, 1, n + 2, 1, a[i] - 1) + 1;
dp[0][i][1] = max(one.query(1, 1, n + 2, a[i] + 1, n + 2) + 1, dp[0][i][0]);
zero.update(1, 1, n + 2, a[i], dp[0][i][0]);
one.update(1, 1, n + 2, a[i], dp[0][i][1]);
}
zero.init(); one.init();
for(int i = n; i; --i) {
dp[1][i][0] = zero.query(1, 1, n + 2, 1, a[i] - 1) + 1;
dp[1][i][1] = max(one.query(1, 1, n + 2, a[i] + 1, n + 2) + 1, dp[1][i][0]);
zero.update(1, 1, n + 2, a[i], dp[1][i][0]);
one.update(1, 1, n + 2, a[i], dp[1][i][1]);
}
for(int i = 1; i <= n; ++i) {
len = max(len, dp[0][i][0] + dp[1][i][1] - 1);
}
for(int i = 1; i <= n; ++i) {
assert(len >= dp[0][i][1] + dp[1][i][0] - 1);
}
for(int i = 1; i <= n; ++i) {
if(dp[0][i][0] + dp[1][i][1] - 1 == len) {
up[dp[0][i][0]].push_back(i);
}
if(dp[0][i][1] + dp[1][i][0] - 1 == len) {
down[dp[1][i][0]].push_back(i);
}
}
int cur = 0, flag = 0;
for(int i = 1; i <= len; ++i) {
if(!flag) {
vector<int> &v = up[i];
int p = upper_bound(v.begin(), v.end(), cur) - v.begin();
if(p == v.size() || a[v[p]] <= a[cur]) {
flag = 1;
}
else {
vector<int> &vv = down[len - i + 1];
int pp = upper_bound(v.begin(), v.end(), cur) - v.begin();
if(pp != vv.size() && a[vv[pp]] < a[cur] && vv[pp] < v[p]) {
flag = 1;
}
else {
cur = v[p];
ans[0].push_back(v[p]);
}
}
}
if(flag) {
vector<int> &v = down[len - i + 1];
int p = upper_bound(v.begin(), v.end(), cur) - v.begin();
assert(a[v[p]] < a[cur]);
assert(p < v.size());
cur = v[p];
ans[0].push_back(v[p]);
}
}
cur = flag = 0;
for(int i = 1; i <= len; ++i) {
if(!flag) {
vector<int> &v = up[i];
int p = lower_bound(v.begin(), v.end(), cur, [&](int i, int j){return a[i] > a[j];}) - v.begin() - 1;
if(p < 0 || a[v[p]] <= a[cur]) {
flag = 1;
}
else {
cur = v[p];
ans[1].push_back(v[p]);
}
}
if(flag) {
vector<int> &v = down[len - i + 1];
int p = lower_bound(v.begin(), v.end(), cur, [&](int i, int j){return a[i] < a[j];}) - v.begin() - 1;
assert(p >= 0);
assert(a[v[p]] < a[cur]);
cur = v[p];
ans[1].push_back(v[p]);
}
}
for(int _ = 0; _ < 2; ++_) {
for(int i = 0; i < len; ++i)
printf("%d%c", ans[_][i], " \n"[i + 1 == len]);
}
}
return 0;
}
1005. Everything Is Generated In Equal Probability
solved at 01:45(+1)
求当前序列的逆序对数并且将当前序列随机替换成一个子序列(\(2^n\))种，递归处理每次结果相加直到序列为空为止
每次给定一个\({\rm N}\), 你随机取一个\(n\in[1,{\rm N}]\)，然后随机取一个\(n\)个数的排列作为初始序列，求总逆序对数的期望
设\(f(n)\)代表长为\(n\)的序列的期望
显然$$f(n) = \frac {n(n-1)} 4 + \sum\limits_{i = 0}^{n}\frac {C_n^i} {2^n}f(i)$$
\(n\)只有\(3000\),\(n^2\)递推即可，最后再求个前缀和
群里有老哥一眼看出答案是\(\frac {n^2-1} 9\) ....
1006. Fantastic Magic Cube
upsolved
题意转化之后就是你有一个\(n*n*n\)的立方体，每个\(1*1*1\)的小立方体都有自己的权值（权值为三个坐标的异或），你每次沿着一个方向把一个立方体切成两个立方体，获得的价值是两个立方体的权值和的乘积，你需要把它们都切成\(1*1*1\)的，求最大价值
听了dls题解，发现是个阅读理解，不管怎么切都是一样的
考虑两个不同的小立方体，它们的权值乘积一定被加到了最终的价值里，并且只加了一次
于是只要\(FWT\)求出每个权值的小立方体有多少个然后求一求和就好了
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int mod = 998244353, N = 1048576 + 10, g = 3;
LL qp(LL a, LL n) {
LL res = 1;
while(n) {
if(n & 1)
res = res * a % mod;
a = a * a % mod;
n >>= 1;
}
return res;
}
const int inv2 = qp(2, mod - 2);
void FWT(int a[], int n) {
for(int d = 1; d < n; d <<= 1)
for(int m = d << 1, i = 0; i < n; i += m)
for(int j = 0; j < d; ++j) {
int x = a[i + j], y = a[i + j + d];
a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod;
}
}
void IFWT(int a[], int n) {
for(int d = 1; d < n; d <<= 1)
for(int m = d << 1, i = 0; i < n; i += m)
for(int j = 0; j < d; ++j) {
int x = a[i + j], y = a[i + j + d];
a[i + j] = 1LL * (x + y) * inv2 % mod, a[i + j + d] = (1LL * (x - y) * inv2 % mod + mod) % mod;
}
}
int n, a[N], m;
LL sum, ans;
int main() {
while(~scanf("%d", &n)) {
m = n;
sum = ans = 0;
for(; m & (m - 1); m += m & -m);
for(int i = 0; i < m; ++i)
a[i] = 0;
for(int i = 0; i < n; ++i)
a[i] = 1;
FWT(a, m);
for(int i = 0; i < m; ++i)
a[i] = 1LL * a[i] * a[i] % mod * a[i] % mod;
IFWT(a, m);
for(int i = 0; i < m; ++i)
sum = (sum + 1LL * i * a[i]) % mod;
for(int i = 0; i < m; ++i)
ans = (ans + (1LL * i * a[i] % mod * (sum - i) % mod + mod) % mod) % mod;
printf("%lld\n", ans * inv2 % mod);
}
return 0;
}
1009. I Love Palindrome String
待补, PAM待学
1010. Just Skip The Problem
solved at 00:31
直接输出\(n!\)就好了，\(1\)特判掉（不过我看有的人没特判就过了，怕不是没这数据。。。）
1011. Keen on Everything But Triangle
solved at 00:44
\(Q\)次询问一个区间里的数能构成的最大周长的三角形
因为数据范围是\(1e9\)所以大概四十几个数就一定能构成三角形，从大到小排序，先找\(123\)，不行就\(234\)，找个四十几次就行了
主席树实现的，线段树维护前\(45\)大暴力合并据说也可以
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int MAXN = 100010;
struct node {
int l, r, sum;
}Tree[MAXN * 30];
int root[MAXN];
struct num {
int x, id;
bool operator < (const num &rhs) const {
return x < rhs.x;
}
}a[MAXN];
int rnk[MAXN];
int n, m, i, j, k, cnt, l, r, x, y, z;
void init() {
cnt = 0;
root[0] = 0;
Tree[0].l = Tree[0].r = Tree[0].sum = 0;
}
void update(int &rt, int l, int r, int num) {
Tree[++cnt] = Tree[rt];
rt = cnt;
Tree[rt].sum++;
if(l == r)
return;
int mid = l + r >> 1;
if(num <= mid)
update(Tree[rt].l, l, mid, num);
else
update(Tree[rt].r, mid + 1, r, num);
}
int query(int i, int j, int k, int l, int r) {
if(l == r)
return l;
int mid = l + r >> 1;
int d = Tree[Tree[j].l].sum - Tree[Tree[i].l].sum;
if(k <= d)
return query(Tree[i].l, Tree[j].l, k, l, mid);
else
return query(Tree[i].r, Tree[j].r, k - d, mid + 1, r);
}
int main() {
while(~scanf("%d %d", &n, &m)) {
init();
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i].x);
a[i].id = i;
}
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; ++i) {
rnk[a[i].id] = i;
}
for(int i = 1; i <= n; ++i) {
root[i] = root[i - 1];
update(root[i], 1, n, rnk[i]);
}
while(m--) {
scanf("%d %d", &l, &r);
if(r - l < 2) {
printf("-1\n");
continue;
}
cnt = r - l + 1;
x = a[query(root[l - 1], root[r], cnt, 1, n)].x;
y = a[query(root[l - 1], root[r], cnt - 1, 1, n)].x;
z = a[query(root[l - 1], root[r], cnt - 2, 1, n)].x;
if(x < y + z) {
printf("%lld\n", 0LL + x + y + z);
continue;
}
bool flag = 0;
for(int i = 3; i < cnt; ++i) {
x = y;
y = z;
z = a[query(root[l - 1], root[r], cnt - i, 1, n)].x;
if(x < y + z) {
flag = 1;
printf("%lld\n", 0LL + x + y + z);
break;
}
}
if(!flag) {
printf("-1\n");
}
}
}
return 0;
}
1012. Longest Subarray
upsolved
自闭题
求最长的连续子序列满足序列里出现的数都至少出现了\(k\)次
枚举右端点，线段树维护不合法的区间，每次询问最左边的合法位置即可
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int mn[N << 2], lazy[N << 2];
void pushdown(int rt) {
if(lazy[rt]) {
lazy[rt << 1] += lazy[rt];
lazy[rt << 1 | 1] += lazy[rt];
mn[rt << 1] += lazy[rt];
mn[rt << 1 | 1] += lazy[rt];
lazy[rt] = 0;
}
}
void pushup(int rt) {
mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);
}
void build(int rt, int l, int r) {
lazy[rt] = mn[rt] = 0;
if(l == r)
return;
int mid = l + r >> 1;
build(rt << 1, l, mid);
build(rt << 1 | 1, mid + 1, r);
}
void update(int rt, int l, int r, int L, int R, int val) {
if(L <= l && r <= R) {
mn[rt] += val;
lazy[rt] += val;
return;
}
pushdown(rt);
int mid = l + r >> 1;
if(L <= mid)
update(rt << 1, l, mid, L, R, val);
if(R > mid)
update(rt << 1 | 1, mid + 1, r, L, R, val);
pushup(rt);
}
int query(int rt, int l, int r) {
if(mn[rt] > 0) return 0;
if(l == r) return l;
pushdown(rt);
int mid = l + r >> 1;
if(mn[rt << 1] == 0)
return query(rt << 1, l, mid);
return query(rt << 1 | 1, mid + 1, r);
}
int a[N], n, c, k, ans, i, j, x, y;
vector<int> pos[N];
int main() {
while(~scanf("%d%d%d", &n, &c, &k)) {
if(k == 1) {
for(int i = 1; i <= n; ++i)
scanf("%*d");
printf("%d\n", n);
continue;
}
build(1, 1, n);
ans = 0;
for(i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
if(pos[a[i]].empty()) {
update(1, 1, n, 1, i, 1);
pos[a[i]].push_back(i);
}
else {
x = pos[a[i]].size() - k;
y = pos[a[i]].back();
if(x < 0) {
update(1, 1, n, 1, y, -1);
}
else {
update(1, 1, n, pos[a[i]][x] + 1, y, -1);
}
pos[a[i]].push_back(i);
x = pos[a[i]].size() - k;
y = pos[a[i]].back();
if(x < 0) {
update(1, 1, n, 1, y, 1);
}
else {
update(1, 1, n, pos[a[i]][x] + 1, y, 1);
}
}
j = query(1, 1, n);
if(j) ans = max(ans, i - j + 1);
}
for(int i = 1; i <= c; ++i)
pos[i].clear();
printf("%d\n", ans);
}
return 0;
}
1007dls发了ppt, 估计是学不会，1008队友补了，1009看什么时候有时间去学
浙公网安备 33010602011771号