XVIII Open Cup named after E.V. Pankratiev. Grand Prix of SPb

A. Base $i - 1$ Notation

两个性质：

• $2=1100$
• $122=0$

利用这两条性质实现高精度加法即可。

时间复杂度$O(n)$。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
const int N = 6e6 + 10;
const int inf = 1e9;
int casenum, casei;
char a[N], b[N];
int c[N];
int main()
{
	freopen("base-i-1.in","r",stdin); freopen("base-i-1.out","w",stdout);
	while(~scanf("%s%s", a, b))
	{
		int n = strlen(a); reverse(a, a + n);
		int m = strlen(b); reverse(b, b + m);
		//int n = 5e5; for(int i = 0; i < n; ++i)a[i] = '1'; a[n] = 0;
		//int m = 5e5; for(int i = 0; i < m; ++i)b[i] = '1'; b[m] = 0;
		int g = max(n, m);
		for(int i = 0; i < g; ++i)
		{
			int p = i / 2;

			c[i] = 0;
			if(i < n)c[i] += a[i] - 48;
			if(i < m)c[i] += b[i] - 48;
		}
		for(int i = 0; i < g; ++i)
		{
			int t=min(c[i]/2,min(c[i+1]/2,c[i+2]));
			c[i]-=t*2;
			c[i+1]-=t*2;
			c[i+2]-=t;
			int w = c[i] / 2;
			if(w)
			{
				c[i + 2] += w;
				c[i + 3] += w;
				g = max(g, i + 4);
			}
			c[i] %= 2;
		}
		while(g>1&&!c[g-1])g--;
		for(int i = g - 1; i >= 0; --i)printf("%d", c[i]);
		puts("");
	}
	return 0;
}
/*

*/

　　

B. Squaring a Bit

直接暴力枚举平方数即可通过，需要手写popcount。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
const int N = 3e5 + 10;
typedef long long LL;
LL n;
int v[65555];
int popcount(LL x){
  return v[x&65535]+v[x>>16&65535]+v[x>>32&65535]+v[x>>48];
}
int main()
{
    for(int i=1;i<65536;i++)v[i]=v[i>>1]+(i&1);
	freopen("bit-squares.in","r",stdin); freopen("bit-squares.out","w",stdout);
	while(~scanf("%lld", &n))
	{
		LL len = 0;
		int one = 0;
		while(n)
		{
			++len;
			one += n & 1;
			n /= 2;
		}

		LL bott = 1ll << (len - 1);
		int bot = sqrt(bott) + 0.99999999;
		LL topp = (1ll << len) - 1;
		int top = sqrt(min(topp, (LL)1e18));
		//printf("%lld %lld\n", bott, topp);
		//printf("%d %d\n", bot, top);
		int ans = 0;
		LL val = (LL)bot * bot, add = bot * 2 + 1;
		int i=bot;
		for(;i+4<=top;i+=4){
            if(popcount(val) == one)++ans;
            val += add, add += 2;
            if(popcount(val) == one)++ans;
            val += add, add += 2;
            if(popcount(val) == one)++ans;
            val += add, add += 2;
            if(popcount(val) == one)++ans;
            val += add, add += 2;
		}
		for(; i <= top; ++i, val += add, add += 2)
		{
			//printf("%d %lld\n", i, val);
			//val = (LL)i * i;
			if(popcount(val) == one)++ans;
		}
		printf("%d\n", ans);
	}
	return 0;
}
/*

*/

　　

C. Chickens

状压DP求方案数。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
const int N = 3e5 + 10;
const int inf = 1e9;
int casenum, casei;
int c[15], e[15];
int f[15][1 << 13];
int n;
int main()
{
	freopen("chickens.in","r",stdin);
	freopen("chickens.out","w",stdout);
	while(~scanf("%d", &n))
	{
		for(int i = 0; i < n; ++i)scanf("%d", &c[i]);
		for(int i = 0; i < n; ++i)scanf("%d", &e[i]);
		memset(f, 0, sizeof(f)); f[0][0] = 1;
		int top = (1 << n) - 1;
		for(int i = 0; i < n; ++i)
		{
			for(int j = 0; j <= top; ++j)if(f[i][j])
			{
				for(int k = 0; k < n; ++k)if((~j >> k & 1) && c[i] <= e[k])
				{
					f[i + 1][j | 1 << k] += f[i][j];
				}
			}
		}
		printf("%d\n", f[n][top]);
	}
	return 0;
}
/*

*/

　　

D. Lights at a Crossing

对于每个观察，枚举两盏红灯，由剩余时间大的向小的连权值为差值的有向边，则最小周期为这个有向图的最小环。

时间复杂度$O(n^3+mn^2)$。

#include<cstdio>
const int N=30,inf=10000000;
int n,m,i,j,k,f[N][N],a[N],ans=inf;char s[99];
inline void up(int&x,int y){x>y?(x=y):0;}
int main(){
  scanf("%d%d",&n,&m);
  for(i=1;i<=n;i++)for(j=1;j<=n;j++)f[i][j]=inf;
  while(m--){
    for(i=1;i<=n;i++){
      scanf("%s",s);
      if(s[0]=='X')a[i]=-1;else sscanf(s,"%d",&a[i]);
    }
    for(i=1;i<=n;i++)if(~a[i])for(j=i+1;j<=n;j++)if(~a[j]){
      if(a[i]<a[j])up(f[i][j],a[j]-a[i]);
      if(a[i]>a[j])up(f[j][i],a[i]-a[j]);
    }
  }
  for(k=1;k<=n;k++)for(i=1;i<=n;i++)for(j=1;j<=n;j++)up(f[i][j],f[i][k]+f[k][j]);
  for(i=1;i<=n;i++)up(ans,f[i][i]);
  if(ans==inf)ans=-1;
  printf("%d",ans);
}

　　

E. Decimal Form

法雷序列求两分数之间分母最小的分数。

时间复杂度$O(T\log w)$。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef __int128 lll;
typedef pair<lll,lll>P;
lll n,r,a,b,c,d,t;
int Case;
ld goal;
const int lim=1000000000;
const ld eps=1e-15;
lll gcd(lll a,lll b){return b?gcd(b,a%b):a;}
P cal(lll a,lll b,lll c,lll d){
	lll x=a/b+1;
	if(x*d<c)return P(x,1);
	if(!a)return P(1,d/c+1);
	if(a<=b&&c<=d){
		P t=cal(d,c,b,a);
		swap(t.first,t.second);
		return t;
	}
	x=a/b;
	P t=cal(a-b*x,b,c-d*x,d);
	t.first+=t.second*x;
	return t;
}
void write(lll x){
	if(x>=10)write(x/10);
	x%=10;
	printf("%d",(int)x);
}
//0.666666666666666667
int main(){
	freopen("decimal-form.in","r",stdin);
	freopen("decimal-form.out","w",stdout);
	scanf("%d",&Case);
	while(Case--){
		n=18;
		static char s[100];
		scanf("%s",s);
		int len=strlen(s);
		r=0;
		for(int i=2;i<len;i++)r=r*10+s[i]-'0';
		if(!r){
			puts("0 1");
			continue;
		}
		for(t=10;n--;t*=10);
		a=r*10-5,b=t;
		c=r*10+5,d=t;
		lll o=gcd(a,b);
		a/=o,b/=o;
		o=gcd(c,d);
		c/=o,d/=o;
		P p=cal(a,b,c,d);
		if(b<p.second)p=P(a,b);
		write(p.first);
		putchar(' ');
		write(p.second);
		puts("");
	}
}

　　

F. Martian Maze

留坑。

G. Wet Mole

Floodfill求出所有水能灌到的点即可。

时间复杂度$O(nm)$。

#include<cstdio>
const int N=1010;
int n,m,i,j,flag;char a[N][N];bool v[N][N];
inline bool check(int x,int y){
	if(x<1||x>n||y<1||y>m)return 0;
	if(a[x][y]=='#')return 0;
	return 1;
}
void dfs(int x,int y){
	if(v[x][y])return;
	v[x][y]=1;
	if(check(x+1,y))dfs(x+1,y);
	else{
		if(check(x,y-1))dfs(x,y-1);
		if(check(x,y+1))dfs(x,y+1);
	}
}
int main(){
	freopen("mole.in","r",stdin);
	freopen("mole.out","w",stdout);
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)scanf("%s",a[i]+1);
	for(i=1;i<=m;i++)if(a[1][i]=='.')dfs(1,i);
	for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(a[i][j]=='.'&&!v[i][j]){
		if(!flag){
			flag=1;
			a[i][j]='X';
		}
	}
	if(flag){
		puts("Yes");
		for(i=1;i<=n;i++)puts(a[i]+1);
	}else puts("No");
}

　　

H. Oddities

留坑。

I. Sorting on the Plane

将所有向量分成两类：在$1$号向量左侧和右侧的。然后分别排序即可。

时间复杂度$O(n\log n)$。

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int N=100010;
int n,i,f[N],m,t;
vector<int>q[2];
int ask(int x,int y){
	printf("? %d %d\n",x,y);
	fflush(stdout);
	int t;
	scanf("%d",&t);
	return t;
}
bool cmp(int x,int y){return ask(x,y);}
int main(){
	scanf("%d",&n);
	if(n==1){
		puts("! YES");
		puts("1");
		fflush(stdout);
		return 0;
	}
	for(i=2;i<=n;i++){
		q[ask(1,i)].push_back(i);
	}
	for(i=0;i<2;i++)sort(q[i].begin(),q[i].end(),cmp);
	for(i=0;i<q[0].size();i++)f[++m]=q[0][i];
	f[++m]=1;
	for(i=0;i<q[1].size();i++)f[++m]=q[1][i];
	if(t=ask(f[1],f[m]))puts("! YES");else puts("! NO");
	if(!t)printf("%d ",m);
	for(i=1;i<=m;i++)printf("%d ",f[i]);
	fflush(stdout);
	return 0;
}

　　

J. Center of List of Sums

二分求出上下界以及上下界的排名，然后将中间的数暴力取出即可。

时间复杂度$O(n\log n)$。

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2000010;
int n,m,i;ll a[N],b[N],q[N],L,R,down,up,cntdown,cntup;
inline ll cal(ll lim){//<=lim
	int i=1,j=n;
	ll t=0;
	for(i=1;i<=n;i++){
		while(j&&a[i]+b[j]>lim)j--;
		t+=j;
	}
	return t;
}
inline ll getkth(ll k){
	ll l=0,r=2100000000,mid,t;
	while(l<=r){
		mid=(l+r)>>1;
		if(cal(mid)>=k)r=(t=mid)-1;else l=mid+1;
	}
	return t;
}
void push(ll A,ll B){
	int i,j=n,k=n;
	for(i=1;i<=n;i++){
		while(j&&a[i]+b[j]>=A)j--;
		while(k&&a[i]+b[k]>B)k--;
		for(int o=j+1;o<=k;o++)q[++m]=a[i]+b[o];
	}
}
int main(){
	freopen("sums-center.in","r",stdin);
	freopen("sums-center.out","w",stdout);
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
	for(i=1;i<=n;i++)scanf("%lld",&b[i]);
	sort(a+1,a+n+1);
	sort(b+1,b+n+1);
	L=1LL*n*(n-1)/2+1;
	R=1LL*n*(n+1)/2;
	down=getkth(L);
	up=getkth(R);
	//printf("L=%lld R=%lld down=%lld up=%lld\n",L,R,down,up);
	if(down==up){
		for(ll o=L;o<=R;o++)printf("%lld ",down);
		return 0;
	}
	cntdown=cal(down);//<=down
	cntup=cal(up-1);//<up
	//printf("cnt=%lld %lld\n",cntdown,cntup);
	push(down+1,up-1);
	for(ll o=L;o<=cntdown&&o<=R;o++)q[++m]=down;
	for(ll o=R;o>cntup&&o>=L;o--)q[++m]=up;
	sort(q+1,q+m+1);
	for(i=1;i<=m;i++)printf("%lld ",q[i]);
}

　　

K. Cookies

将每个串排序，然后直接DP求出最长链即可。

#include<cstdio>
#include<string>
#include<algorithm>
#include<map>
#include<iostream>
using namespace std;
const int N=180000;
int n,m,i;
int len[N];
int f[N],g[N];
string name[N],show[N],need[N];
map<string,int>pos;
string q[N];
int dp(int x){
	if(!x)return 0;
	if(f[x])return f[x];
	int A=f[x],B=0;
	for(int i=0;i<name[x].size();i++){
		string now=name[x];
		now.erase(i,1);
		int o=pos[now];
		if(!o)continue;
		int t=dp(o);
		if(t>A)A=t,B=o;
	}
	f[x]=A+1,g[x]=B;
	return f[x];
}
int main(){
	freopen("word-chains.in","r",stdin); freopen("word-chains.out","w",stdout);
	cin>>n;//n=10000
	for(i=0;i<n;i++)cin>>need[i];
	cin>>m;//173554
	m++;
	name[1]="";
	show[1]=".";
	for(i=2;i<=m;i++){
		cin>>name[i];
		show[i]=name[i];
	}
	for(i=1;i<=m;i++){
		sort(name[i].begin(),name[i].end());
		pos[name[i]]=i;//sorted
		len[i]=name[i].size();
	}
	for(i=0;i<n;i++){
		string now=need[i];
		sort(now.begin(),now.end());
		int x=pos[now];
		dp(x);
		int cnt=0;
		while(x){
			q[++cnt]=show[x];
			x=g[x];
		}
		q[1]=need[i];
		cout<<cnt<<endl;
		for(int j=1;j<=cnt;j++){
			cout<<q[j];
			if(j<cnt)cout<<" -> ";else cout<<endl;
		}
	}
}
/*
4
university
open
cup
cookie
17
university
intrusive
neuritis
unities
seniti
nisei
sine
sei
e
es
one
ne
open
cup
up
p
cookie
*/

　　

L. Xor-fair Division

若所有数异或和非$0$，则答案为$0$，否则答案为$2^n-2$。

#include<cstdio>
typedef long long ll;
int n;long long ans,sum,x;
int main(){
	freopen("xorseq.in","r",stdin);
	freopen("xorseq.out","w",stdout);
	scanf("%d",&n);
	ans=1LL<<n;
	while(n--){
		scanf("%lld",&x);
		sum^=x;
	}
	if(sum)ans=0;
	else if(n==1)ans=0;
	else ans-=2;
	printf("%lld",ans);
}