# HGOI 20190828 题解

Problem A 数学题

Solution : 我们思考对于$r-l \leq 10^7$怎么处理，

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100005;
ll a[505] = {0,16504974832918ll,49514807041302ll,82524641299076ll,115534479934505ll,148544254405234ll,181554158308885ll,214563872594611ll,247573764207188ll,280583538103016ll,313593361626057ll,346602992946568ll,379613110929775ll,412622718262607ll,445632601916252ll,478642437755379ll,511652194457614ll,544661947496041ll,577671637940336ll,610681752050814ll,643691404588574ll,676701244176651ll,709711054775594ll,742720800286221ll,775730381310093ll,808740585896190ll,841750243027296ll,874760117073666ll,907769696410853ll,940779525226536ll,973789738206002ll,1006799160955474ll,1039808877324861ll,1072819018026025ll,1105828761150350ll,1138838541727837ll,1171848079131282ll,1204858104382139ll,1237867536035320ll,1270877842610423ll,1303887672241912ll,1336897155489979ll,1369907175171416ll,1402916647740933ll,1435927028696641ll,1468936230981939ll,1501946419251818ll,1534955574922012ll,1567966183816364ll,1600975595444924ll,1633985443056901ll,1666995678674159ll,1700005364180681ll,1733014763642307ll,1766024840505475ll,1799034631307151ll,1832044002952252ll,1865054030679522ll,1898064068452051ll,1931073520776903ll,1964083598976891ll,1997093098588907ll,2030103003059564ll,2063112702217827ll,2096122737310551ll,2129133054850411ll,2162142198328595ll,2195152115983905ll,2228161264785474ll,2261172544708734ll,2294181378777450ll,2327191233379424ll,2360201373720367ll,2393211269318476ll,2426220486575075ll,2459230819150224ll,2492240046752302ll,2525251064525340ll,2558259845985451ll,2591268636654841ll,2624280077930602ll,2657288734859772ll,2690299477550963ll,2723308150991146ll,2756319304183776ll,2789328790852636ll,2822338626083272ll,2855347891381860ll,2888357969641819ll,2921367932058355ll,2954377671380517ll,2987387687179118ll,3020395919679682ll,3053407972625776ll,3086417613231907ll,3119426240525339ll,3152436403285179ll,3185446530826116ll,3218455004607940ll,3251466542904717ll,3284474102416981ll,3317485456042933ll,3350494835559564ll,3383505323561767ll,3416514694107384ll,3449524616389762ll,3482535136132263ll,3515544067001265ll,3548552960601121ll,3581563404252177ll,3614573730318868ll,3647583278380671ll,3680593224979870ll,3713603049026829ll,3746614512962914ll,3779623637711295ll,3812631445796912ll,3845642667207059ll,3878650989452908ll,3911661656239387ll,3944671816767424ll,3977681760271560ll,4010691144633474ll,4043702388095629ll,4076710028369395ll,4109720346259831ll,4142730350170032ll,4175739772832184ll,4208750686895433ll,4241759731456393ll,4274769641755841ll,4307777872223390ll,4340789892971559ll,4373799166736030ll,4406809512953890ll,4439817532702298ll,4472829307111790ll,4505836643327275ll,4538848618830486ll,4571857797045020ll,4604867900156142ll,4637878508493697ll,4670886796634050ll,4703896752766576ll,4736906975045792ll,4769917026147829ll,4802926090607880ll,4835937748578967ll,4868944057292098ll,4901955388347787ll,4934964803726201ll,4967976054354879ll,5000985142582223ll,5033994769269421ll,5067004519698299ll,5100015475381224ll,5133023523525477ll,5166035548680638ll,5199044102202040ll,5232053769509974ll,5265063976719498ll,5298071256992732ll,5331086032306110ll,5364093009886904ll,5397101286590264ll,5430112363775912ll,5463125358224685ll,5496130321891579ll,5529143222333432ll,5562151266937027ll,5595161916107909ll,5628170955930451ll,5661180935315217ll,5694191463640877ll,5727200867537917ll,5760210082013019ll,5793220944569250ll,5826231266751643ll,5859240047889761ll,5892248661408006ll,5925259731715720ll,5958268524424498ll,5991280302232593ll,6024290558811088ll,6057297413541092ll,6090308767576979ll,6123319740301532ll,6156326709064649ll,6189337531550686ll,6222350666231978ll,6255356723766425ll,6288367605824763ll,6321377226929040ll,6354387848727584ll,6387396476006881ll,6420406295019013ll,6453418361894346ll,6486425279806916ll,6519434933117769ll,6552447537063411ll,6585455288301656ll,6618465571313108ll,6651473219530721ll,6684487597609615ll,6717493272112836ll,6750503897921972ll,6783514370819588ll,6816524779187087ll,6849533841224964ll,6882543200522966ll,6915555567048002ll,6948563176458473ll,6981569435885717ll,7014587121946570ll,7047593188567860ll,7080603174566593ll,7113612266982300ll,7146622391827726ll,7179631332633195ll,7212642629393380ll,7245652465198424ll,7278661503882743ll,7311671230143188ll,7344681060552161ll,7377691956846305ll,7410698365443030ll,7443713861235034ll,7476716796858509ll,7509729721125427ll,7542738638892030ll,7575747728275743ll,7608761023356292ll,7641769632981677ll,7674779747704999ll,7707788455568150ll,7740801037220385ll,7773805775344854ll,7806820992857333ll,7839827911666137ll,7872836137053706ll,7905848048044634ll,7938859068637500ll,7971867868233332ll,8004875132181171ll,8037889423443128ll,8070896110232830ll,8103906084599451ll,8136918977662318ll,8169925767524876ll,8202934185046077ll,8235946072038061ll,8268954361637275ll,8301967510589725ll,8334974643904362ll,8367982557462832ll,8400994813545019ll,8434004890087843ll,8467013576718783ll,8500020663978651ll,8533037879707351ll,8566043768970011ll,8599053931450993ll,8632062998388608ll,8665072097208580ll,8698085691961056ll,87310937660715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ll l, r;
ll _div[10000005];
bool nt_prime[N];
int prime_cnt;
int prime[N];
void get_prime(int n) {
for (int i = 2; i <= n; i++) {
if (!nt_prime[i]) prime[++ prime_cnt] = i;
for (int j = 1; j <= prime_cnt && prime[j] * i <= n; j++) {
nt_prime[i * prime[j]] = 1;
if (i % prime[j] == 0) break;
}
}
}
ll calc(ll l, ll r) {
memset(_div, 0, sizeof _div);
ll res = 0;
for (ll i = 1; i <= prime_cnt; i++) {
for (ll j = (l - 1) / prime[i] * prime[i] + prime[i]; j <= r; j += prime[i]) {
if (!_div[j - l]) _div[j - l] = prime[i];
}
}
for (ll i = l; i <= r; i++) {
if (_div[i - l] && _div[i - l] != i) res += i / _div[i - l];
}
return res;
}
ll solve(ll x) {
ll res = 0;
ll w = x / 10000000;
for (int i = 1; i <= w; i++) res += a[i];
res += calc(w * 10000000 + 1, x);
return res;
}
int main() {
get_prime(1e5);
scanf("%lld%lld", &l, &r);
printf("%lld\n", solve(r) - solve(l - 1));
return 0;
}
A.cpp

Problem B 假题

Solution： 直接从$1$跑树的深度，贪心，深度大的优先，用一次dfs除去当前新加入点集的点附近$L$的点。

对于随机数据，复杂度应该是$O(能过)$

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N=5e5+10;
struct rec{
int pre,to,w;
}a[N<<1];
bool vis[N];
bool cmp(int a,int b){return d[a]>d[b];}
{
int X=0,w=0; char c=0;
while(c<'0'||c>'9') {w|=c=='-';c=getchar();}
while(c>='0'&&c<='9') X=(X<<3)+(X<<1)+(c^48),c=getchar();
return w?-X:X;
}
{
a[tot].to=v;
a[tot].w=w;
}
void dfs1(int u,int fa)
{
int v=a[i].to; if (v==fa) continue;
d[v]=d[u]+a[i].w; dfs1(v,u);
}
}
void dfs2(int u,int fa,int L)
{
if (L<=0) return;
vis[u]=1;
int v=a[i].to; if (v==fa) continue;
dfs2(v,u,L-a[i].w);
}
}
signed main()
{
for (int i=1;i<=n-1;i++) {
}
dfs1(1,0);
for (int i=1;i<=n;i++) p[i]=i;
sort(p+1,p+1+n,cmp);
int ans = 0;
for (int i=1;i<=n;i++) if (!vis[p[i]]) dfs2(p[i],0,l),ans++;
printf("%lld\n",ans);
return 0;
}
B.cpp

Problem C 计数题

1. 对于任意$1 \leq i \leq n$,都有$b_i | b_{i+n}$且$b_{i+n} | a_i$

2. $\prod\limits_{i=n+1}^{2n} b_i \leq \prod\limits_{i=1}^{n} {b_i}^2$

Solution :

显然，划分阶段的时候，$i$和$i+n$是一个整体，不可分割。

如果$a_i$足够小，我们可以直接记$f[i][j]$表示当前放到第$i$个和第$i+n$个b序列后半部分乘积除去b序列前半部分的乘积的商为$j$的方案数。

于是这样暴力的状态和dp，是一个$O(n \prod a_i \sqrt{a_i})$的美妙算法，空间复杂度到了阶乘级别。

显然，这样状态数就爆炸了，我们不可能将累乘所获得的值记到状态中去，然后我们会思考怎么不将累乘记到附加状态去。

很快，我们会发现根本没办法做，这个状态不得不记。

敏锐的观察到有一个$a_i = 2^k$的部分分，因子只有一个！附加状态显然可以用对数优化到$log_2 a_i$级别。

由于因数的对称性，即以$\sqrt{a_i}$为对称轴，大于$\sqrt{a_i}$和小于$\sqrt{a_i}$的因数个数都是一样的。

由于选择的数，都是独立的，互不干扰，那么满足$\prod\limits_{i=n+1}^{2n} b_i \leq \prod\limits_{i=1}^{n} {b_i}^2$和满足$\prod\limits_{i=n+1}^{2n} b_i \geq \prod\limits_{i=1}^{n} {b_i}^2$的序列个数都是一样的。

对于不考虑第二个限制的方案数，比较容易求，就是$\prod \frac{(c+1)(c+2)}{2}$ ，其中$c$表示某个数，某一个因数出现的次数。

于是，我们需要考虑恰好，$\prod\limits_{i=n+1}^{2n} b_i = \prod\limits_{i=1}^{n} {b_i}^2$的方案数。

对于每一个独立的因数，我们都可以按照$2$那样处理(对数记做状态)，如果当前数不是其倍数，那么就直接跳过，继续下面的转移。

最后将每个独立的质因子的答案相乘就是最后的答案。

设$ans1$表示不考虑第二种方案限制的序列个数，$ans2$表示满足$\prod\limits_{i=n+1}^{2n} b_i = \prod\limits_{i=1}^{n} {b_i}^2$的序列个数。

最终的答案就是$\frac{ans1+ans2}{2}$

具体的动态规划方程还是比较经典的也没有什么需要特别注意的地方。

复杂度应该是$O(Prime\_Num \times n \times \sum_{i=1}^{n} log a_i \times {log_2}^2 Max\{a_i\}))$

# include <bits/stdc++.h>
# define int long long
using namespace std;
const int N=105,mo=998244353;
int f[N][6001],n,a[N],rec[N],t[N];
bool is_pr[31623];
int pr[31623];
vector<int>v;
int Pow(int x,int n) {
int ans = 1;
while (n) {
if (n&1) ans=ans*x%mo;
x=x*x%mo;
n>>=1;
}
return ans%mo;
}
void EouLaShai(int Lim)
{
memset(is_pr,true,sizeof(is_pr));
is_pr[1]=false;
for (int i=2;i<=Lim;i++) {
if (is_pr[i]) pr[++pr[0]]=i;
for (int j=1;j<=pr[0]&&i*pr[j]<=Lim;j++) {
is_pr[i*pr[j]]=false;
if (i%pr[j]==0) break;
}
}
}
signed main() {
int inv2=Pow(2,mo-2); EouLaShai(31622);
int ret1=1;
scanf("%lld",&n);
for (int i=1;i<=n;i++) scanf("%lld",&a[i]);
memcpy(rec,a,sizeof(a));
for (int i=1;i<=n;i++) {
for (int j=1;j<=pr[0];j++) {
if (pr[j]>rec[i]) break;
if (rec[i]%pr[j]==0) {
v.push_back(pr[j]); int c=0;
while (rec[i]>1 && rec[i]%pr[j]==0) rec[i]/=pr[j],c++;
ret1=ret1*((c+1)*(c+2)/2)%mo;
}
}
if (rec[i]!=1) ret1=ret1*((1+1)*(1+2)/2)%mo,v.push_back(rec[i]);
}
sort(v.begin(),v.end());
v.erase(unique(v.begin(),v.end()),v.end());
int ret2=1;
for (int tmp=0;tmp<v.size();tmp++) {
int num = v[tmp],sum=0;
for (int i=1;i<=n;i++) {
t[i]=0; int mht=a[i];
while (mht>1&&mht%num==0) mht/=num,t[i]++;
sum+=t[i];
}
for (int i=0;i<=n;i++)
for (int j=-sum;j<=sum;j++)
f[i][j+3000]=0;
f[0][0+3000]=1;
for (int i=0;i<=n-1;i++)
for (int j=-sum;j<=sum;j++) if (f[i][j+3000]) {
if (a[i+1]%num!=0) { f[i+1][j+3000]=f[i][j+3000]; continue;}
int lim = t[i+1];
for (int k=0;k<=lim;k++)
for (int w=0;w<=k;w++) if (j+k-2*w>=-sum &&j+k-2*w<=sum) {
f[i+1][j+k-2*w+3000]=(f[i+1][j+k-2*w+3000]+f[i][j+3000])%mo;
}
}
ret2=ret2*f[n][0+3000]%mo;
}
int ans = (ret1+ret2)*inv2%mo;
printf("%lld\n",ans);
return 0;
}
C.cpp

posted @ 2019-08-28 23:22  ljc20020730  阅读(144)  评论(0编辑  收藏  举报