10.5 模拟赛
\(\rm NOIP\) 模拟赛
\(\rm Date: 10.5\)
上次真是太变态了，这次就平和许多
另外，不要在意accoders那丧心病狂的评测机速度了
今天4道都是CF原题
\(T1\) CF1252G
一看上去居然没思路啊，于是就把T2做了，再回来看这题终于发现了做法。其实我们只用维护当前有多少人的能力值比自己小，如果太小了就寄了。拿线段树维护区间加区间最小即可。
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "inline", "-ffast-math")
#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#define rg register
#define emp emplace_back
using namespace std;
const int N = 1e5 + 5;
namespace Sg {
#define lsp p << 1
#define rsp p << 1 | 1
#define mid (t[p].l + t[p].r >> 1)
struct T {
int l, r, val; // Minimum
}t[N << 2];
int tag[N << 2]; // Num to add
inline T operator + (T L, T R) {
return {L.l, R.r, min(L.val, R.val)};
}
inline void pushup(int p) { t[p].val = min(t[lsp].val, t[rsp].val); }
inline void fx(int p, int v) {
t[p].val += v, tag[p] += v;
}
inline void pushdown(int p) {
fx(lsp, tag[p]), fx(rsp, tag[p]), tag[p] = 0;
}
void build(int *arr, int p, int l, int r) {
t[p] = {l, r}; if(l == r) return t[p].val = arr[l], void();
build(arr, lsp, l, mid), build(arr, rsp, mid + 1, r); pushup(p);
}
void update(rg int p, rg int l, rg int r, rg int v) {
if(l <= t[p].l && t[p].r <= r) return fx(p, v); pushdown(p);
if(l <= mid) update(lsp, l, r, v); if(mid < r) update(rsp, l, r, v); pushup(p);
}
int query(int p, int l, int r) {
if(l <= t[p].l && t[p].r <= r) return t[p].val; pushdown(p);
if(r <= mid) return query(lsp, l, r); else if(l > mid) return query(rsp, l, r);
else return min(query(lsp, l, r), query(rsp, l, r));
}
}
int n, m, q, kk;
int b[N], c[N];
vector <int> a[N];
signed main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("Ans.txt", "w", stdout);
#endif
scanf("%d %d %d", &n, &m, &q);
for(int i = 0; i <= m; ++i) a[i].emp(0);
for(int i = 1, x; i <= n; ++i) {
scanf("%d", &x), a[0].emp(x);
if(i == 1) kk = x;
else {
if(x < kk) ++b[0];
}
}
//cout << kk << endl;
c[0] = n;
for(int i = 1, num; i <= m; ++i) {
scanf("%d", &num);
c[i] = num;
for(int j = 1, x; j <= num; ++j) {
scanf("%d", &x);
a[i].emp(x);
if(x < kk) ++b[i];
}
}
//for(int i = 0; i <= m; ++i) cout <<
//for(int i = 0; i <= m; ++i) cout << b[i] << ' ';
//cout << endl;
for(int i = 1; i <= m; ++i) b[i] = b[i - 1] - c[i] + b[i];
for(int i = 0; i < m; ++i) b[i] -= c[i + 1];
//for(int i = 0; i <= m; ++i) cout << b[i] << ' ';
//cout << endl;
Sg::build(b, 1, 0, m);
for(int x, y, z; q--; ) {
scanf("%d %d %d", &x, &y, &z);
int val = a[x][y];
//cout << a[x][y] << endl;
if(val < kk && z > kk) Sg::update(1, x, m, -1);
if(val > kk && z < kk) Sg::update(1, x, m, 1);
if(Sg::t[1].val < 0) puts("0");
else puts("1");
a[x][y] = z;
//printf(": %d\n", Sg::query(1, 2, 2));
}
return 0;
}
\(T2\) CF1515E
网上到处是 \(\mathcal{O(n^2)}\) 的做法还一个比一个奇怪，解释都不带解释的直接上柿子。哎，还是自己看得懂自己的解法
显然两端极长手开的电脑之间只会隔一个电脑，那么设 \(f[i, j]\) 表示考虑前 \(i\) 个电脑分 \(j\) 段手开，那么有转移：
\[f[i, j] = (\sum\limits_{x=1}^{i-2}f[i-x-1][j-1])\times C_{i-j+1}^x \times 2^{x-1}
\]
\(\mathcal{O(n^3)}\)，accoders上 \(100\to 25\)
#include <bits/stdc++.h>
#pragma GCC opimtize("Ofast")
#define rg register
using namespace std;
const int N = 405;
using LL = long long;
int n, mod;
// first i computers, j blocks -> total ways
int f[N][N];
LL jc[N], fjc[N], pw[N];
LL power(LL x, int y) {
if(!y) return 1;
LL t = power(x, y >> 1);
return t * t % mod * (y & 1 ? x : 1) % mod;
}
inline void init() {
jc[0] = pw[0] = 1;
for(int i = 1; i <= 400; ++i) jc[i] = 1LL * jc[i - 1] * i % mod;
for(int i = 1; i <= 400; ++i) fjc[i] = power(jc[i], mod - 2);
for(int i = 1; i <= 400; ++i) pw[i] = 2LL * pw[i - 1] % mod;
}
inline LL C(int x, int y) { // C(x, y) (x >= y)
return jc[x] * fjc[y] % mod * fjc[x - y] % mod;
}
int g[N][N];
inline LL C_(int x, int y) {
return g[x][y];
}
signed main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("Ans.txt", "w", stdout);
#endif
cin >> n >> mod;
init();
//printf("%d\n", C(5, 3));
f[1][1] = 1;
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= n; ++j) {
g[i][j] = C(i, j);
}
}
for(rg int i = 2; i <= n; ++i) {
f[i][1] = pw[i - 1];
for(rg int j = 2; j <= n; ++j) {
for(rg int x = 1; x < i - 1; ++x) {
LL tmp = C_(i - j + 1, x) * pw[x - 1] % mod;
//printf("i = %d, j = %d, tmp = %lld\n", i, j, tmp);
//printf("from [%d, %d]\n", i - x - 1, j - 1);
(f[i][j] += f[i - x - 1][j - 1] * tmp % mod) %= mod;
//f[i][j] = (f[i][j] + )
//f[i][j] = (f[i - x - 1][j - 1] + tmp) % mod;
}
}
}
#if 0
for(int i = 1; i <= n; ++i) {
for(int j = 1; j <= n; ++j) {
printf("f[%d, %d] = %d\n", i, j, f[i][j]);
}
}
#endif
LL ans = 0;
for(int j = 1; j <= n; ++j) (ans += f[n][j]) %= mod;
cout << ans;
//cout << endl << clock();
return 0;
}
\(T3\) CF1654F
当时发现一条性质就是 \(i\to i\ \text{xor}\ e\) 的时候 \(s_i\) 是有规律的交换的，然后luogu的题解就用了一个类似SA倍增的思路解决本题。
另一种思路就是预处理从 \(i\) 开始，变换 \(2^j\) 次的构成的字符串的哈希值（巧妙地利用了 \(i\to i\text{ xor }e\) 时“下一个字符”的变换规律），然后我们判断字典序就利用哈希值找到第一个不相同的字符的位置，从而判断两个生成串的字典序。
\(\mathcal{O(n\cdot2^n)}\)
#include <bits/stdc++.h>
#define rg register
#define BRUTE_FORCE 0
#define SOL 9
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 3e5 + 5;
#if BRUTE_FORCE
string ret = "";
inline string Trans(string s, int val) {
string ret = "";
for(rg int i = 0; i < n; ++i) {
ret += s[i ^ val];
//cout << (i ^ val) << ' ';
}
//cout << endl;
return ret;
}
inline char find(string s) {
char ret = 'z';
for(rg int i = 0; i < n; ++i) ret = min(ret, s[i]);
return ret;
}
inline string solve(string s) {
ret = Trans(s, 0);
char ch = find(s);
for(rg int i = 1; i < n; ++i) {
if(s[i] != ch) continue;
string tmp = Trans(s, i);
if(ret > tmp) ret = tmp;
}
return ret;
}
#endif
#if SOL
using ULL = unsigned long long;
const ULL C[] = {998244353, ULL(1e9 + 7)};
ULL f[2][20][N];
ULL pw[2][N];
int n;
char s[N];
int cmp(int x, int y) {
if(f[0][n][x] == f[0][n][y] && f[1][n][x] == f[1][n][y]) return 0;
for(int j = n - 1; ~j; --j)
if(f[0][j][x] == f[0][j][y] && f[1][j][x] == f[1][j][y])
x xor_eq 1 << j, y xor_eq 1 << j;
return s[x] < s[y] ? -1 : 1;
}
void work() {
cin >> n >> s;
pw[0][0] = pw[1][0] = 1;
for(int j = 0; j < 2; ++j)
for(int i = 1; i < (1 << n); ++i) pw[j][i] = pw[j][i - 1] * C[j];
for(int i = 0; i < (1 << n); ++i) f[0][0][i] = f[1][0][i] = s[i] * s[i];
for(int k = 0; k < 2; ++k) {
for(int j = 1; j <= n; ++j) {
for(int i = 0; i < (1 << n); ++i)
f[k][j][i] = f[k][j - 1][i] + f[k][j - 1][i ^ (1 << (j - 1))] * pw[k][1 << (j - 1)];
}
}
int ans = 0;
for(int i = 1; i < (1 << n); ++i)
if(cmp(ans, i) > 0) ans = i;
for(int i = 0; i < (1 << n); ++i) putchar(s[ans ^ i]);
return;
}
#endif
signed main() {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("Ans.txt", "w", stdout);
#endif
work();
}
\(T4\) CF1710D
*3400 诶 有人当场A了orzycy
大概思路就是从小到大考虑每个极长连通块，他必然包含了若干个子连通块区间。
设我们考虑到了好区间 \([l,r]\)，\(l\) 所在连通块为 \([l',x_1
]\)，\(r\) 所在连通块为 \([x_k+1,r']\)，它们之间还有 \(k-1\) 个连通块 \([x_i+1,x_{i+1}]\)。满足 $l'\le l\le x_1<...<x_k\le r \le r'l $
且 \(k\ge 1\)。
容易发现 \(k=2\) 时无解。\(k=1\) 时直接连 \((l,r)\) 即可。\(k>2\) 时一种构造是连 \((l,r),(l,x_k)\) 以及 \((x_i,r)(2\le i<k)\)。
不写了
#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 5;
int n, T, flg = 0;
int fa[N];
char s[N][N];
signed work() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) fa[i] = i, scanf("%s", s[i] + i);
for(int i = 1; i <= n; ++i)
for(int j = i - 1; j >= 1; --j)
if(s[j][i] == '1' && fa[i] > j) {
printf("%d %d\n", j, i);
if(fa[fa[i] - 1] > j) {
printf("%d %d\n", j, fa[i] - 1);
for(int k = fa[fa[i] - 1] - 1; k >= j + 1; --k)
if(fa[k] == k) printf("%d %d\n", k, i);
}
for(int k = j; k <= i; ++k) fa[k] = fa[j];
}
return 0;
}
int main() {
if(!T && !flg) {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("Ans.txt", "w", stdout);
#endif
cin >> T, flg = 1;
}
if(!T && flg) return 0;
work();
--T;
return main();
}
