XVI Open Cup named after E.V. Pankratiev. GP of Peterhof

A. (a, b)-Tower

当指数大于模数的时候用欧拉定理递归计算,否则直接暴力计算。

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<iostream>
using namespace std;
typedef long long LL;
int pw(int x,int y,int mod){
	int ret=1;
	for(int i=0;i<y;i++){
		ret=1LL*ret*x%mod;
	}
	return ret;
}
int solve(int l,int r,int p){
	if(l==1)return 1;
	if(l==r){
		return l%p;
	}
	int phi=0;
	for(int i=1;i<=p;i++)if(__gcd(p,i)==1)phi++;
	return pw(l,solve(l+1,r,phi)+phi,p);
}
int main(){
	freopen("abtower.in","r",stdin);
	freopen("abtower.out","w",stdout);
	int a,b;
	while(scanf("%d%d",&a,&b)!=EOF){
		printf("%d\n",solve(a,b,10));
	}
}

  

B. Bridges Construction

留坑。

 

C. Equivalence Relation

留坑。

 

D. Formula-1

留坑。

 

E. Ideal Photo

三分第一个人的位置即可。

#include <bits/stdc++.h>
using namespace std ;

const int MAXN = 205 ;
int n;
vector<int>V;
double check(double x){
	double ret=0;
	for(int i=0;i<V.size();i++){
		ret+=sqrt(1.0+1.0*(V[i]-x-i)*(V[i]-x-i));
	}
	return ret;
}
void solve () {
	V.clear();
	for(int i=1;i<=n;i++){
		int x;scanf("%d",&x);
		V.push_back(x);
	}
	for(int i=1;i<=n;i++){
		int x;scanf("%d",&x);
		V.push_back(x);
	}
	sort(V.begin(),V.end());
	double l=-3e6;
	double r=3e6;
	for(int i=0;i<300;i++){
		double l1=l+(r-l)/3,l2=l+(r-l)/3*2;
		double t1=check(l1),t2=check(l2);
		if(t1>t2)l=l1;else r=l2;
	}
	double ans=check(l);
	//printf("%.12f %.12f\n",l,r);
	printf("%.12f\n",ans);
	//for(double l=-5;l<=5;l+=0.1)printf("%.12f\n",check(l));
}

int main () {
	freopen ( "make-a-row.in" , "r" , stdin ) ;
	freopen ( "make-a-row.out" , "w" , stdout ) ;
	while ( ~scanf ( "%d" , &n ) ) solve () ;
	return 0 ;
}

  

F. (p, q)-Knight

首先通过exgcd求出一组可行解,然后暴力调整若干项即可。

#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<time.h>
#include<assert.h>
#include<iostream>
using namespace std;
typedef long long LL;
typedef pair<int,int>pi;
const LL Inf=1LL<<60;
LL exgcd(LL a,LL b,LL &x,LL &y){
	if(!b)return x=1,y=0,a;
	LL d=exgcd(b,a%b,y,x);
	y-=a/b*x;return d;
}
LL ans;
void up(LL &x,LL y){if(x>y)x=y;}
void check(LL t1,LL t2,LL x,LL y){
	if(abs(t1%2)!=abs(x%2))return;
	if(abs(t2%2)!=abs(y%2))return;
	up(ans,max(abs(t1),abs(x))+max(abs(t2),abs(y)));
	//if(x==10&&y==-3)printf("val=%lld\n",max(abs(t1),abs(x))+max(abs(t2),abs(y)));
	//if(x==10&&y==-3)printf("t1=%lld t2=%lld ans=%lld\n",t1,t2,ans);
}
int main(){
	freopen("pqknight.in","r",stdin);
	freopen("pqknight.out","w",stdout);
	LL p,q;
	while(scanf("%lld%lld",&p,&q)!=EOF){
		if(p>q)swap(p,q);
		if(p==0){
			puts(q==1?"1":"-1");
			continue;
		}		
		if(p==1){
			if(q%2==1)puts("-1");
			else printf("%lld\n",q+1);
			continue;
		}
		if(__gcd(p,q)!=1){
			puts("-1");
			continue;
		}
		LL x,y;
		LL t1=p,t2=q;
		exgcd(p,q,x,y);
		//printf("x=%lld y=%lld\n",t1,t2);
		ans=Inf;
		for(int i=-2;i<=2;i++){
			check(t1,t2,x+i*q,y-i*p);
		}
		x=-x,y=-y;
		for(int i=-2;i<=2;i++){
			check(t1,t2,x+i*q,y-i*p);
		}
		if(ans==Inf){
			puts("-1");
		}
		else printf("%lld\n",ans);
	}	
	return 0;
}

  

G. Random Wormholes

如果能到1,那么直接走过去,否则随机走向下一个点。

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<iostream>
using namespace std;
typedef long long LL;

int main(){
	//freopen("abtower.in","r",stdin);
	//freopen("abtower.out","w",stdout);
	while(1){
		LL RAND=(1LL<<32)-1;
		printf("1\n");
		fflush(stdout);
		string s;
		cin>>s;
		if(s=="yes"){break;}
		while(1){
			LL x=(rand()%65536)*65536LL+rand()%65536;
			printf("%lld\n",x);
			fflush(stdout);
			cin>>s;
			if(s=="yes")break;
		}
	}
}

  

H. Second Maximum

分段积分。

#include <bits/stdc++.h>
using namespace std ;

const int MAXN = 205 ;
int n;
vector<int>V;
int l[55],r[55];
double a[55];
double b[55];
void mul(double t0,double t1){
	memset(b,0,sizeof b);
	for(int i=0;i<=n+1;i++){
		if(i)b[i]+=a[i-1]*t1;
		b[i]+=a[i]*t0;
	}
	for(int i=0;i<=n+1;i++)a[i]=b[i];
}
void solve () {
	double ans=0;
	V.clear();
	for(int i=0;i<n;i++){
		scanf("%d%d",l+i,r+i);
		V.push_back(l[i]);
		V.push_back(r[i]);
	}
	sort(V.begin(),V.end());
	V.erase(unique(V.begin(),V.end()),V.end());
	for(int it=0;it<V.size()-1;it++){
		int curl=V[it],curr=V[it+1];
		for(int i=0;i<n;i++){
			if(r[i]<=curl)continue;
			memset(a,0,sizeof a);
			a[0]=1;
			//if(curr>l[i])mul((curr+0.0)/(r[i]-l[i]),1.0/(l[i]-r[i]));
			bool flag=1;
			for(int j=0;j<n;j++){
				if(i==j)continue;
				if(curr<=l[j]){flag=0;break;}
				if(curl<r[j]){
					mul((l[j]+0.0)/(l[j]-r[j]),1.0/(r[j]-l[j]));
				}
			}
			if(!flag)continue;
			/*
			if(it==1&&i==0){
				puts("wa");
				for(int j=0;j<=n;j++)printf("%.2f ",a[j]);puts("");
			}
			*/
			a[0]=0;
			for(int j=1;j<=n;j++)a[j]*=j;
			if(curr>l[i])mul((r[i]+0.0)/(r[i]-l[i]),1.0/(l[i]-r[i]));
			for(int j=1;j<=n+1;j++)a[j]/=j+1;
			double rr=curr*curr,ll=curl*curl;
			for(int j=1;j<=n+1;j++){
				ans+=a[j]*(rr-ll);
				rr=rr*curr;
				ll=ll*curl;
			}
			//printf("it=%d i=%d ans=%.2f\n",it,i,ans);
		}
	}
	printf("%.12f\n",ans);
}
int main(){
	freopen("secondmax.in","r",stdin);
	freopen("secondmax.out","w",stdout);
	while(~scanf("%d",&n))solve();
}

  

I. Ticket To Ride

留坑。

 


总结:

  • 尽快适应数学场。

 

posted @ 2016-11-16 20:20 Claris 阅读(...) 评论(...) 编辑 收藏