北京一零一中2024年信息学迎新马拉松题解
A T469715 [2024迎新马拉松] 101
相当于选择一段长度为 \(3k\) 的区间使得变化的总值最小。维护每一个元素变化到 \(1\) 与 \(0\) 的要求数量，之后前缀和处理即可。
#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
const ll MAXN=1e6+5;
ll t0[MAXN],t1[MAXN];
ll v0(ll l,ll r){
return t0[r]-t0[l-1];
}
ll v1(ll l,ll r){
return t1[r]-t1[l-1];
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
ll n,k;
cin>>n>>k;
for(int i=1;i<=n;++i){
ll a;
cin>>a;
t0[i]+=t0[i-1]+min(10-a,a);
t1[i]+=t1[i-1]+min(11-a,(a==0?1:a-1));
}
ll val=1e18;
for(int i=1;i<=n-3*k+1;++i){
val=min(val,v1(i,i+k-1)+v0(i+k,i+2*k-1)+v1(i+2*k,i+3*k-1));
}
cout<<n-3*k+val<<endl;
return 0;
}
B T467650 [2024迎新马拉松] 排列
设 \(f_{i,j}\) 表示直到第 \(i\) 个变换时 \(x\) 要变到 \(j\) 最多的保留数量。
则有
- \(f_{i,j}=f_{i-1,j}+1 (j\neq a_i,j\neq b_i)\)
- \(f_{i,a_i}=\max(f_{i-1,a_i},f_{i-1,b_i}+1)\)
- \(f_{i,b_i}=\max(f_{i-1,b_i},f_{i-1,a_i}+1)\)
线段树优化动规解决。
#include <bits/stdc++.h>
#define endl "\n"
#define lc(u) u<<1
#define rc(u) u<<1|1
using namespace std;
typedef long long ll;
const ll MAXN=2e6+5;
ll n,m,x;
ll t[MAXN*4],tag[MAXN*4];
void push_down(ll u,ll l,ll r){
ll mid=(l+r)>>1;
t[lc(u)]+=(mid-l+1)*tag[u];
tag[lc(u)]+=tag[u];
t[rc(u)]+=(r-mid)*tag[u];
tag[rc(u)]+=tag[u];
tag[u]=0;
}
ll query(ll u,ll l,ll r,ll x){
if(l==r){
return t[u];
}
push_down(u,l,r);
ll mid=(l+r)>>1;
if(x<=mid){
return query(lc(u),l,mid,x);
}
return query(rc(u),mid+1,r,x);
}
void change(ll u,ll l,ll r,ll x,ll val){
if(l==r){
t[u]=val;
return;
}
push_down(u,l,r);
ll mid=(l+r)>>1;
if(x<=mid){
change(lc(u),l,mid,x,val);
return;
}
change(rc(u),mid+1,r,x,val);
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>n>>m>>x;
memset(t,~0x3f,sizeof(t));
change(1,1,n,x,0);
for(int i=1;i<=m;++i){
ll a,b;
cin>>a>>b;
ll va=query(1,1,n,a),vb=query(1,1,n,b);
tag[1]++;
change(1,1,n,a,max(va,vb+1));
change(1,1,n,b,max(vb,va+1));
}
for(int i=1;i<=n;++i){
cout<<(query(1,1,n,i)<0?-1:query(1,1,n,i))<<" ";
}
return 0;
}
C
D T469844 [2024迎新马拉松] 质数
线性筛筛出所有 \([1,10^6]\) 内的质数然后枚举每一个点开始的子串累计即可。
#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
const ll MAXN=1e7+5;
ll p[MAXN],tot;
bool np[MAXN];
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
for(int i=2;i<1000000;++i){
if(!np[i]){
p[++tot]=i;
}
for(int j=1;j<=tot&&i*p[j]<1000000;++j){
np[i*p[j]]=true;
if(i%p[j]==0){
break;
}
}
}
ll k;
cin>>k;
string s;
cin>>s;
s=" "+s;
ll n=s.size()-1,ans=0;
np[1]=true;
for(int i=1;i<=n-k+1;++i){
if(s[i]=='0'){
continue;
}
ll val=0;
for(int j=i;j<=i+k-1;++j){
val=val*10+s[j]-'0';
}
ans+=!np[val];
}
cout<<ans<<endl;
return 0;
}
E T469864 [2024迎新马拉松] 无限串
用vector维护每一个字母的所有出现位置，然后根据循环把两端的单独二分查找有几个数，中间的除一下看看能分出几段，拼合即可，但是需要高精度。
#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
const ll MAXN=5e6;
const ll MOD=1e9+7;
unsigned to[55][MAXN];
ll s[MAXN];
ll query(ll x,ll l,ll r){
return to[x][r]-to[x][l-1];
}
ll ksm(ll x,ll b){
ll ans=1;
while(b){
if(b&1){
ans=ans*x%MOD;
}
x=x*x%MOD;
b>>=1;
}
return ans;
}
ll k[MAXN],o[MAXN];
string ms(string a,ll b){
ll sz=0;
for(int i=a.size()-1;i>=0;--i){
k[++sz]=a[i]-'0';
}
ll i=0;
while(b){
++i;
k[i]-=b%10;
if(k[i]<0){
k[i]+=10;
k[i+1]-=1;
}
b/=10;
}
string c;
bool good=false;
for(i=sz;i>=1;--i){
if(good||k[i]!=0){
good=true;
c+=char(k[i]+'0');
}
}
return c;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
string S,t;
cin>>S>>t;
bool isd=false;
for(int i=0;i<t.size();++i){
if(isdigit(t[i])){
isd=true;
break;
}
}
ll n=S.size(),m=t.size();
S=" "+S;
t=" "+t;
for(int i=1;i<=n;++i){
if('a'<=S[i]&&S[i]<='z'){
s[i]=S[i]-'a'+1;
}else{
s[i]=S[i]-'A'+27;
}
}
for(int i=1;i<=n;++i){
for(int j=1;j<=53;++j){
to[j][i]=to[j][i-1];
}
to[s[i]][i]++;
}
ll ans=0;
ll id=0;
for(int i=1;i<=m;++i){
ll c;
if('a'<=t[i]&&t[i]<='z'){
c=t[i]-'a'+1;
}else{
c=t[i]-'A'+27;
}
ll tot=0;
string num;
if(isdigit(t[i+1])){
while(i+1<=m&&isdigit(t[i+1])){
num+=t[i+1];
++i;
}
}
if(num.empty()){
num="1";
}
if(num.size()<=10){
ll v=0;
for(auto c:num){
v=v*10+c-'0';
}
if(v>query(c,id+1,n)){
v-=query(c,id+1,n);
ans+=n-id;
id=0;
ans+=((v)/query(c,1,n))*n;
v-=((v)/query(c,1,n))*query(c,1,n);
}
if(v==0){
ans-=n;
v+=query(c,1,n);
}
ll nid=lower_bound(to[c]+id+1,to[c]+n+1,to[c][id]+v)-to[c];
ans+=nid-id;
id=nid;
if(id==n){
id=0;
}
}else{
num=ms(num,query(c,id+1,n));
ans+=n-id;
ans%=MOD;
string nv=num;
vector<ll>basic;
ll left=0;
for(auto ca:nv){
left=left*10+ca-'0';
basic.push_back(left/query(c,1,n));
left%=query(c,1,n);
}
ll b=1;
for(int j=basic.size()-1;j>=0;--j){
ans+=basic[j]*b%MOD*n%MOD;
b*=10;
b%=MOD;
}
if(left==0){
ans=((ans-n)%MOD+MOD)%MOD;
left+=query(c,1,n);
}
id=0;
ll nid=lower_bound(to[c]+id+1,to[c]+n+1,to[c][id]+left)-to[c];
ans+=nid-id;
id=nid;
if(id==n){
id=0;
}
}
}
cout<<unsigned(ans%MOD)<<endl;
return 0;
}
F T467532 [2024迎新马拉松] 字典
先把所有东西离线下来处理好。
之后把字典的字符串按字典序排序，然后求出每个询问对应前缀的字符串的区间 \([l,r]\)，显然这个区间里的所有元素的前缀都是一样的。那么考虑后缀。
按顺序将每个元素倒过来加进可持久化字典树中，之后只要查询 \([l,r]\) 范围内的后缀元素数量即可。
#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;
const ll MAXN=1e6+5;
string s[MAXN];
map<string,ll>l,r;
set<string>sl,sr;
ll n,m;
struct query{
string f,b;
}q[MAXN];
ll ans[MAXN];
ll trie[MAXN*4][26],tot;
ll sum[MAXN*4],rt[MAXN*4];
void insert(ll &u,string s){
++tot;
for(int i='a';i<='z';++i){
trie[tot][i-'a']=trie[u][i-'a'];
}
sum[tot]=sum[u]+1;
u=tot;
ll v=u;
for(int i=0;i<s.size();++i){
ll x=s[i]-'a';
++tot;
ll on=trie[v][x];
for(int j=0;j<26;++j){
trie[tot][j]=trie[on][j];
}
sum[tot]=sum[on]+1;
trie[v][x]=tot;
v=trie[v][x];
}
}
ll query(ll lu,ll ru,string s){
for(int i=0;i<s.size();++i){
ll c=s[i]-'a';
lu=trie[lu][c];
ru=trie[ru][c];
}
return sum[ru]-sum[lu];
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>n>>m;
for(int i=1;i<=n;++i){
cin>>s[i];
}
sort(s+1,s+n+1);
for(int i=1;i<=m;++i){
string c;
cin>>c;
int j=0;
for(j=0;c[j]!='*';++j){
q[i].f+=c[j];
}
j++;
while(j<c.size()){
q[i].b+=c[j];
++j;
}
reverse(q[i].b.begin(),q[i].b.end());
sl.insert(q[i].f);
}
for(ll i=1;i<=n;++i){
string val;
if(sl.count(val)){
if(!l.count(val)){
l[val]=i;
}
r[val]=i;
}
for(int j=0;j<s[i].size();++j){
val+=s[i][j];
if(sl.count(val)){
if(!l.count(val)){
l[val]=i;
}
r[val]=i;
}
}
reverse(s[i].begin(),s[i].end());
rt[i]=rt[i-1];
insert(rt[i],s[i]);
}
for(int i=1;i<=m;++i){
cout<<query(rt[l[q[i].f]-1],rt[r[q[i].f]],q[i].b)<<endl;
}
return 0;
}
G T467526 [2024迎新马拉松] 回路
考虑二分中位数，对于二分的 \(x\)，将数组的元素抽象成：
- \(a_{i,j}=1(a_{i,j}\geq x)\)
- \(a_{i,j}=-1(a_{i,j}<x)\)
- \(a_{1,1}=0\)
然后我们要做的是在这个数组上找一条最大的回路使得这条回路的值大于 \(0\)。考虑费用流连边即可。
#include <bits/stdc++.h>
#define endl "\n"
#define int long long
using namespace std;
typedef long long ll;
const ll MAXN=300;
const ll MAXM=MAXN*MAXN;
const ll INF=1e18;
ll s, t;
struct edge {
ll to, weight, cost, nxt;
} e[MAXN * MAXN* 20];
ll head[MAXN*MAXN*2], tot = 1;
void add(ll u, ll v, ll w, ll c) {
e[++tot]={v,w,c,head[u]};
head[u] = tot;
e[++tot]={u,0,-c,head[v]};
head[v]=tot;
}
ll cost;
ll dis[MAXN*MAXN*2];
bool inq[MAXN*MAXN*2];
struct path {
ll edge, pre;
} p[MAXN*MAXN*2];
void clear(){
memset(head,0,sizeof(head));
tot=1;
cost=0;
}
bool spfa() {
memset(inq, false, sizeof(inq));
for (int i = 0; i < MAXN*MAXN*2; ++i) {
dis[i] = INF;
}
queue<ll> q;
q.push(s);
inq[s] = true;
dis[s] = 0;
while (!q.empty()) {
ll u = q.front();
q.pop();
inq[u] = false;
for (ll i = head[u]; i; i = e[i].nxt) {
ll v = e[i].to, c = e[i].cost, w = e[i].weight;
if (dis[u] + c < dis[v] && w) {
dis[v] = dis[u] + c;
p[v].edge = i;
p[v].pre = u;
if (!inq[v]) {
inq[v] = true;
q.push(v);
}
}
}
}
return dis[t] != INF;
}
ll ek() {
ll ans = 0;
while (spfa()) {
ll Min = INF;
for (ll i = t; i != s; i = p[i].pre) {
Min = min(Min, e[p[i].edge].weight);
}
ans += Min;
for (ll i = t; i != s; i = p[i].pre) {
e[p[i].edge].weight -= Min;
e[p[i].edge ^ 1].weight += Min;
}
cost += Min * dis[t];
}
return ans;
}
ll n,m,a[MAXN][MAXN];
ll b[MAXN][MAXN];
ll gidi(ll x,ll y){
return (x-1)*m+y;
}
ll gido(ll x,ll y){
return n*m+gidi(x,y);
}
bool check(ll x){
for(int i=1;i<=n;++i){
for(int j=1;j<=m;++j){
if(a[i][j]>=x){
b[i][j]=1;
}else{
b[i][j]=-1;
}
}
}
b[1][1]=0;
clear();
s=0,t=MAXN*MAXN-2000;
add(s,gidi(1,1),2,0);
add(gido(n,m),t,2,0);
for(int i=1;i<=n;++i){
for(int j=1;j<=m;++j){
ll in=gidi(i,j),out=gido(i,j);
if((i==1&&j==1)||(i==n&&j==m)){
add(in,out,2,-b[i][j]);
}else{
add(in,out,1,-b[i][j]);
}
if(j<m){
add(gido(i,j),gidi(i,j+1),1,0);
}
if(i<n){
add(gido(i,j),gidi(i+1,j),1,0);
}
}
}
ll val=ek();
ll ans=cost+b[n][m];
//cout<<x<<" "<<val<<" "<<ans<<endl;
return ans<=0;
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>n>>m;
for(int i=1;i<=n;++i){
for(int j=1;j<=m;++j){
cin>>a[i][j];
}
}
ll ans=0,l=1,r=1e9;
while(l<=r){
ll mid=(l+r)>>1;
if(check(mid)){
l=mid+1;
ans=mid;
}else{
ans=mid-1;
r=mid-1;
}
}
cout<<ans<<endl;
return 0;
}
H
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