XVII Open Cup named after E.V. Pankratiev. GP of SPb
A. Array Factory
将下标按前缀和排序，然后双指针，维护最大的右边界即可。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=200010;
int n,i,j,anslen,ansl,ansr,mr,q[N];
ll a[N],lim;
inline bool cmp(int x,int y){return a[x]<a[y];}
int main(){
freopen("arrayfactory.in","r",stdin);
freopen("arrayfactory.out","w",stdout);
scanf("%d%lld",&n,&lim);
for(i=1;i<=n;i++)scanf("%lld",&a[i]),a[i]+=a[i-1];
for(i=0;i<=n;i++)q[i]=i;
sort(q,q+n+1,cmp);
for(i=j=0;i<=n;i++){
while(j<=n&&a[q[j]]-a[q[i]]<=lim)mr=max(mr,q[j++]);
if(mr>q[i]){
int len=mr-q[i];
if(len>anslen){
anslen=len;
ansl=q[i];
ansr=n-mr;
}else if(len==anslen&&q[i]<ansl){
ansl=q[i];
ansr=n-mr;
}
}
}
if(!anslen)return puts("-1"),0;
printf("%d\n%d %d",anslen,ansl,ansr);
}
/*
a[r]-a[l]<=lim
ask max r
a[r]<=a[l]+lim
*/
B. Purchases and Bonuses
$f[i][j]$表示购买了前$i$个物品，目前有$j$积分时最多省多少钱，转移就是要么直接买，要么把积分全用完。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef pair<LL,LL>pi;
const int Maxn=102,Inf=1e9;
int n;
int a[Maxn];
int pre[102][100020];
int pw[102][100020];
int rep[102];
int dp[2][100020];
int main(){
freopen("bonuses.in","r",stdin);
freopen("bonuses.out","w",stdout);
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=n;i++)scanf("%d",a+i);
int cs=0;
for(int i=0;i<=n*1000;i++)dp[cs][i]=Inf;
dp[cs][0]=0;
for(int i=1;i<=n;i++,cs^=1){
for(int j=0;j<=n*1000;j++)dp[cs^1][j]=Inf;
for(int j=0;j<=i*1000;j++){
if(dp[cs][j]==Inf)continue;
if(dp[cs^1][j+a[i]/100]>dp[cs][j]+a[i]){
dp[cs^1][j+a[i]/100]=dp[cs][j]+a[i];
pre[i][j+a[i]/100]=j;
pw[i][j+a[i]/100]=0;
}
int tmp=min(a[i],j);
int tmpv=dp[cs][j]+a[i]-tmp;
if(dp[cs^1][j-tmp]>tmpv){
dp[cs^1][j-tmp]=tmpv;
pre[i][j-tmp]=j;
pw[i][j-tmp]=tmp;
}
}
}
int ans=Inf,anscs;
for(int i=0;i<=n*1000;i++){
if(ans>dp[cs][i]){
ans=dp[cs][i];anscs=i;
}
}
for(int i=n;i>=1;i--){
rep[i]=pw[i][anscs];
anscs=pre[i][anscs];
}
printf("%d\n",ans);
for(int i=1;i<=n;i++)printf("%d + %d\n",a[i]-rep[i],rep[i]);
}
}
/*
a[r]-a[l]<=lim
ask max r
a[r]<=a[l]+lim
*/
C. Number of Solutions
留坑。
D. Cutting Potatoes
暴力枚举即可。
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef pair<LL,LL>pi;
int n,K;
struct Node{
LL x,y;
Node(){}
Node(LL x,LL y):x(x),y(y){}
bool operator <(const Node&a)const{
return x*a.y<=y*a.x;
}
Node operator /(const Node&a)const{
return Node(x*a.y,y*a.x);
}
};
int cmp(Node a,Node b){
if(a.x*b.y==a.y*b.x)return 0;
return a.x*b.y>a.y*b.x?1:-1;
}
int a[102];
int rep[102],tmprep[102];
int main(){
freopen("cut-potatoes.in","r",stdin);
freopen("cut-potatoes.out","w",stdout);
while(scanf("%d%d",&n,&K)!=EOF){
for(int i=1;i<=n;i++)scanf("%d",a+i);
Node ans=Node(10000000,1);
for(int i=1;i<=n;i++){
for(int j=1;j<=K;j++){
Node minx=Node(a[i],j);
Node maxx=Node(1,10000000);
bool flag=1;
for(int k=1;k<=n;k++){
int x=a[k]*minx.y/minx.x;
if(!x){flag=0;break;}
x=min(x,K);
tmprep[k]=x;
maxx=max(Node(a[k],x),maxx);
}
if(!flag){continue;}
Node tans=maxx/minx;
if(tans<ans){
ans=tans;
for(int k=1;k<=n;k++)rep[k]=tmprep[k];
}
}
}
for(int i=1;i<=n;i++)printf("%d%c",rep[i],i==n?'\n':' ');
}
}
/*
a[r]-a[l]<=lim
ask max r
a[r]<=a[l]+lim
*/
E. Divide and Conquer
DP求出点数为$n$时的划分方案数，然后枚举决策找到第$k$小解即可。
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef vector<int>vi;
struct P{
int x,y;
P(){}
P(int x,int y):x(x),y(y){}
bool operator<(const P&a)const{
if(x!=a.x)return x<a.x;
return y<a.y;
}
P operator -(const P&a)const{
return P(x-a.x,y-a.y);
}
int operator *(const P&a)const{
return x*a.y-y*a.x;
}
}a[33];
int n;
LL Mp[44];
int cal(P t1,P t2,P t3){
int tmp=(t2-t1)*(t3-t1);
return tmp>0;
}
bool cmp(int x,int y){return a[x]<a[y];}
LL dfs(int n){
if(Mp[n]>=0)return Mp[n];
if(!n){return Mp[n]=1;}
LL t=0;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
int l=j-i-1,r=n-2-l;
if((l&1))continue;
LL res1=dfs(l);
LL res2=dfs(r);
t+=res1*res2;
}
}
return Mp[n]=t;
}
vector<P>rep;
void pt(vi cur,LL ned){
if(cur.empty())return;
bool flag=1;
for(int i=0;i<cur.size()&&flag;i++){
for(int j=i+1;j<cur.size()&&flag;j++){
P t1=a[cur[i]],t2=a[cur[j]];
vi vl,vr;
for(int k=0;k<cur.size();k++){
if(k==i||k==j)continue;
if(cal(t1,t2,a[cur[k]]))vl.push_back(cur[k]);
else vr.push_back(cur[k]);
}
if(vl.size()&1)continue;
sort(vl.begin(),vl.end(),cmp);
sort(vr.begin(),vr.end(),cmp);
LL res1=Mp[vl.size()];
LL res2=Mp[vr.size()];
if(ned>=res1*res2)ned-=res1*res2;
else{
rep.push_back(t1);
rep.push_back(t2);
pt(vl,ned/res2);
pt(vr,ned%res2);
flag=0;
break;
}
}
}
}
int main(){
freopen("dnc.in","r",stdin);
freopen("dnc.out","w",stdout);
//cal(P(0,0),P(2,0),P(2,1));
while(scanf("%d",&n)!=EOF){
for(int i=0;i<n;i++){
int x,y;
scanf("%d%d",&x,&y);
a[i]=P(x,y);
}
vi ori(n,0);
for(int i=0;i<n;i++)ori[i]=i;
sort(ori.begin(),ori.end(),cmp);
memset(Mp,-1,sizeof Mp);
dfs(n);
// for(int i=1;i<=n;i++)printf("%lld ",Mp[i]);puts("");
rep.clear();
LL k;scanf("%lld",&k);
pt(ori,k);
for(int i=0;i<rep.size();i++)printf("%d %d\n",rep[i].x,rep[i].y);
}
}
/*
a[r]-a[l]<=lim
ask max r
a[r]<=a[l]+lim
*/
F. Doubling
大范围直接$/2$构造，小范围DP求解。
#include<cstdio>
#include<algorithm>
#include<string>
#include<iostream>
using namespace std;
typedef long long ll;
const int N=200010;
const int inf=100000000;
typedef pair<int,string>P;
int n,i,j;
P f[111];
string cal(int n){
if(n<=100)return f[n].second;
if(n&1)return cal(n-1)+"1";
return "["+cal(n/2)+"]";
}
int main(){
freopen("doubling.in","r",stdin);
freopen("doubling.out","w",stdout);
scanf("%d",&n);
f[0]=P(0,"");
f[1]=P(1,"1");
f[2]=P(2,"11");
for(i=3;i<=100;i++){
f[i]=P(inf,"");
for(j=1;j<i;j++){
P x=f[j];
x.first+=f[i-j].first;
x.second+=f[i-j].second;
f[i]=min(f[i],x);
}
if(i%2==0){
P x(f[i/2].first+2,"["+f[i/2].second+"]");
f[i]=min(f[i],x);
}
}
cout<<cal(n)<<endl;
return 0;
}
/*
a[r]-a[l]<=lim
ask max r
a[r]<=a[l]+lim
*/
G. New Collection
随机100轮算出期望集合大小，然后找到差距最近的即可。
#include <bits/stdc++.h>
using namespace std ;
typedef long long LL ;
typedef pair < int , int > pii ;
#define getid( l , r ) l + r | ( l != r )
set < string> s ;
double magic[7]={10.0,100.0,999.96,6325.61,9517.19,9950.67,9995.01};
void solve () {
string c ;
for ( int i = 1 ; i <= 10000 ; ++ i ) {
cout << "+" << endl ;
fflush ( stdout ) ;
cin >> c ;
s.insert(c);
}
int x = ( int ) s.size () ;
double ret=1e9;
int ans;
for(int i=0;i<7;i++){
double now=fabs(1.0*x-magic[i]);
if(now<ret)ret=now,ans=i;
}
int fin=1;
for(int i=0;i<=ans;i++)fin*=10;
printf("= %d\n",fin);
fflush ( stdout ) ;
}
int main () {
solve () ;
return 0 ;
}
H. Path or Coloring
贪心染色，如果可以$k$染色，那么就好了，否则必然存在长度为$k$的简单路径。
#include<cstdio>
#include<algorithm>
#include<string>
#include<iostream>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int N=1010;
const int M=20010;
const int inf=100000000;
int Case,n,m,K,i,j,x,y,g[N],v[M],nxt[M],ed,vis[N],col[N],pos;
inline void add(int x,int y){v[++ed]=y;nxt[ed]=g[x];g[x]=ed;}
void dfs(int x){
pos++;
for(int i=g[x];i;i=nxt[i])if(col[v[i]])vis[col[v[i]]]=pos;
for(int i=1;;i++)if(vis[i]<pos){
col[x]=i;
break;
}
for(int i=g[x];i;i=nxt[i])if(!col[v[i]])dfs(v[i]);
}
void solve(){
scanf("%d%d%d",&n,&m,&K);
for(ed=0,i=1;i<=n;i++)g[i]=col[i]=0;
while(m--)scanf("%d%d",&x,&y),add(x,y),add(y,x);
for(i=1;i<=n;i++)if(!col[i])dfs(i);
bool flag=0;
for(i=1;i<=n;i++)if(col[i]>K){flag=1;break;}
if(!flag){
printf("coloring");
for(i=1;i<=n;i++)printf(" %d",col[i]);
return;
}
for(i=1;i<=n;i++)if(col[i]==K+1){x=i;break;}
printf("path");
for(i=1;i<=K+1;i++){
printf(" %d",x);
for(j=g[x];j;j=nxt[j])if(col[v[j]]==col[x]-1){
x=v[j];
break;
}
}
}
int main(){
freopen("pathorcoloring.in","r",stdin);
freopen("pathorcoloring.out","w",stdout);
scanf("%d",&Case);
while(Case--){
solve();
puts("");
}
}
I. Double Shuffle
留坑。
J. Timer
按题意模拟。
#include <bits/stdc++.h>
using namespace std ;
typedef long long LL ;
typedef pair < int , int > pii ;
#define clr( a , x ) memset ( a , x , sizeof a )
#define getid( l , r ) l + r | ( l != r )
const int MAXN = 100005 ;
char s[12][65] ;
char a[12][MAXN] ;
char digit[10][8][9] = {
{
".XXXXX..",
"XX..XXX.",
"XX.XXXX.",
"XXXX.XX.",
"XXX..XX.",
"XXX..XX.",
".XXXXX..",
"........",
},
{
"...XX...",
"..XXX...",
".XXXX...",
"...XX...",
"...XX...",
"...XX...",
".XXXXXX.",
"........",
},
{
".XXXXX..",
"XX...XX.",
".....XX.",
"...XXX..",
".XXX....",
"XX......",
"XXXXXXX.",
"........",
},
{
".XXXXX..",
"XX...XX.",
".....XX.",
"..XXXX..",
".....XX.",
"XX...XX.",
".XXXXX..",
"........",
},
{
"...XXX..",
"..XXXX..",
".XX.XX..",
"XX..XX..",
"XXXXXXX.",
"....XX..",
"...XXXX.",
"........",
},
{
"XXXXXXX.",
"XX......",
"XXXXXX..",
".....XX.",
".....XX.",
"XX...XX.",
".XXXXX..",
"........",
},
{
".XXXXX..",
"XX...XX.",
"XX......",
"XXXXXX..",
"XX...XX.",
"XX...XX.",
".XXXXX..",
"........",
},
{
"XXXXXXX.",
"XX...XX.",
"X....XX.",
"....XX..",
"...XX...",
"..XX....",
"..XX....",
"........",
},
{
".XXXXX..",
"XX...XX.",
"XX...XX.",
".XXXXX..",
"XX...XX.",
"XX...XX.",
".XXXXX..",
"........",
},
{
".XXXXX..",
"XX...XX.",
"XX...XX.",
".XXXXXX.",
".....XX.",
"XX...XX.",
".XXXXX..",
"........",
},
} ;
void paint ( int x , int v ) {
for ( int i = 3 ; i < 10 ; ++ i ) {
for ( int j = x ; j < x + 8 ; ++ j ) {
a[i][j] = digit[v][i - 3][j - x] ;
}
}
}
void mark ( int x ) {
a[0][x + 3] = a[0][x + 4] = a[1][x + 3] = a[1][x + 4] = 'X' ;
}
int check ( int x ) {
for ( int i = 0 ; i < 12 ; ++ i ) {
for ( int j = 0 ; j < 60 ; ++ j ) {
if ( s[i][j] == '-' ) continue ;
if ( s[i][j] != a[i][x + j] ) return 0 ;
}
}
return 1 ;
}
void solve () {
for ( int i = 0 ; i < 12 ; ++ i ) {
scanf ( "%s" , s[i] ) ;
}
int ans = 3470 ;
for ( int i = 0 ; ; ++ i ) {
if ( check ( i ) ) {
printf ( "%02d:%02d\n" , ans / 60 , ans % 60 ) ;
return ;
}
ans -= 5 ;
if ( ans < 0 ) ans += 3600 ;
}
}
void show ( int l , int r ) {
for ( int i = 0 ; i < 12 ; ++ i ) {
for ( int j = l ; j < r ; ++ j ) {
printf ( "%c" , a[i][j] ) ;
}
puts ( "" ) ;
}
}
int main () {
clr ( a , '.' ) ;
int x = 0 ;
for ( int i = 0 ; i < 50000 ; i += 12 ) {
if ( x % 5 == 0 ) {
if ( x < 10 ) paint ( i , x ) ;
else {
paint ( i + 4 , x % 10 ) ;
paint ( i - 4 , x / 10 ) ;
}
}
mark ( i ) ;
x = ( x - 1 + 60 ) % 60 ;
}
//show ( 0 , 110 ) ;
solve () ;
/*
for ( int i = 0 ; i < 10 ; ++ i ) {
scanf ( "%d" , &n ) ;
printf ( "{\n" ) ;
for ( int j = 0 ; j < 8 ; ++ j ) {
scanf ( "%s" , tmp ) ;
printf ( "\"%s\",\n", tmp );
}
printf ( "},\n" ) ;
}
*/
return 0 ;
}
K. Ultraprime Numbers
答案最多只有$9$项。
#include<cstdio>
#include<algorithm>
#include<string>
#include<iostream>
using namespace std;
typedef long long ll;
const int N=2010;
const int inf=100000000;
int n,i,j,p[N/10],tot;bool is[N],v[N];
int q[1111],ans;
inline bool check(int x){
if(!is[x])return 0;
static int f[100];
int n=0;
while(x)f[++n]=x%10,x/=10;
for(int i=n;i;i--){
int t=0;
for(int j=i;j;j--){
t=t*10+f[j];
if(!is[t])return 0;
}
}
return 1;
}
int main(){
freopen("ultraprime.in","r",stdin);
freopen("ultraprime.out","w",stdout);
//scanf("%d",&n);
n=2000;
for(i=2;i<=n;i++){
if(!v[i])is[i]=1,p[tot++]=i;
for(j=0;j<tot;j++){
if(i*p[j]>n)break;
v[i*p[j]]=1;
if(i%p[j]==0)break;
}
}
for(i=2;i<=n;i++)if(check(i)){
q[++ans]=i;
// printf("%d %d\n",ans,i);
if(ans==1000)break;
}
scanf("%d",&n);
if(ans<n)puts("-1");else printf("%d",q[n]);
return 0;
}
/*
a[r]-a[l]<=lim
ask max r
a[r]<=a[l]+lim
*/
