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您在这里: front_page » 2020-2021 » teams » namespace » 牛客多校第五场
2020-2021:teams:namespace:牛客多校第五场

这是本文档旧的修订版！

目录

牛客多校第五场

这场的题目好啊！

一般有规律，奇数场可能偏向数学计算，偶数场更考验玛俐码力。在本人的码力实在不足的情况下，看来更喜欢考验思维量的题目。

（成功地首次独立写出了Prim算法/*优先队列优化*/——然而什么也没有发生/*只过了10%样例*/）

D

因为轮换不影响，不妨在纸上按照圆形排序整个序列，找规律会发现其实就是一个简单的最长上升子序列。DP就完事了。

（和弦转位的背景还挺有趣的）

点击以显示 ⇲

点击以隐藏 ⇱ 

#include<stdio.h>
#include<string.h>
 
int j, s, n, t, a[100001], b[100001], ans = 0;
 
int main()
{
    scanf("%d",&n);
    a[0]=-1000000;
    int i;
    for(i=1;i<=n;i++)
	{
        scanf("%d", &b[i]);
        b[i + 2 * n] = b[i + n] = b[i];
    }
    int j;
    for(j = 0; j < n; j++)
	{
	    s = 0;
	    memset(a, 0, sizeof(a));
	    for(i = 0 + j; i < n + j; i++)
		{
	        t = b[i + 1];
	        if(t > a[s]) a[++s] = t;
	        else
			{
	            int l = 1, h = s, m;
	            while(l <= h)
				{
	                m = (l + h) / 2;
	                (t > a[m])?(l = m + 1):(h = m - 1);
	            }
	            a[l] = t;
	        }
	    }
	    ans =(ans>s)?ans:s;
    }
    printf("%d\n",n-ans);
}

E

我们队是用Java过的。本来想写个高精度，奈何不会写，奈何有这样取巧的方法。

我还考虑要不要存入set的时候保证互素，使得高精度只需FFT就行了——后来发现Java竟然可以完美避过。马佬看来之前用过BigInteger类，我还是第一次听说。

为什么不写Python呢？因为编程类的专业课有Java，却没有Python，导致我能看懂与维护Java，但是还完全不会Python……虽然都是面向对象。

（这样看来，C++的面向对象就是zhazha——元首）

点击以显示 ⇲

点击以隐藏 ⇱ 

import java.math.BigInteger;
import java.util.HashSet;
import java.util.Scanner;
 
public class Main
{
    static BigInteger[] a = new BigInteger[100005];
    static boolean[] vis = new boolean[100005];
    int n;
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for (int i = 1; i < n + 1; i++)
        {
            int temp = sc.nextInt();
            a[i] = BigInteger.valueOf(temp);
        }
        sc.close();
        int cnt;
        BigInteger t;
        HashSet<BigInteger> ans = new HashSet<>();
        for (int i = 1; i < n + 1; i++)
        {
            if (vis[i])
            {
                continue;
            }
            vis[i] = true;
            t = BigInteger.valueOf(i);
            cnt = 1;
            while (!a[t.intValue()].equals(BigInteger.valueOf(i)))
            {
                vis[a[t.intValue()].intValue()] = true;
                t = a[t.intValue()];
                cnt++;
            }
            ans.add(BigInteger.valueOf(cnt));
        }
        BigInteger ans1 = BigInteger.valueOf(1);
        for (BigInteger e : ans)
        {
            BigInteger gcd = ans1.gcd(e);
            ans1 = ans1.multiply(e.divide(gcd));
        }
        System.out.println(ans1);
    }
}

看了标程，用C++补一补题。

点击以显示 ⇲

点击以隐藏 ⇱ 

#include<stdio.h>
 
#include<map>
 
using namespace std;
 
int ww;
 
int n, p[100010];
 
map<int,int> fac;
 
int solve(int pos)
{
	int cnt = 0;
	while(p[pos])
	{
		int nxt = p[pos];
		p[pos] = 0;
		pos = nxt;
		++cnt;
	}
	return cnt;
}
 
void factor(int x)
{
	int i;
	for(i = 2; i*i <= x;++i)
	{
		if( x % i == 0 )
		{
			int cnt = 0;
			while(x % i == 0)
			{
				x /= i;
				++cnt;
			}
			fac[i] =max(fac[i], cnt);
		}
	}
	if(x != 1)
	{
		fac[x] =max(fac[x], 1);
	}
}
 
int num[100010],len;
 
void multiply(int x)
{
	int c = 0;
	int i;
	for(i = 0; i < len; ++i)
	{
		long long tmp = (long long)x * num[i] + c;
		num[i] = int(tmp % 1000000);
		c = int(tmp / 1000000);
	}
	if( c )
	{
		num[len++] = c;
	}
}
 
int main()
{
	ww = scanf("%d", &n);
	int i;
	for(i = 1; i <= n; ++i)
	{
		ww = scanf("%d", p+i);
	}
	for(i = 1; i <= n; ++i)
	{
		if(p[i])
		{
			factor(solve(i));
		}
	}
	num[0] = 1, len = 1;
	map<int,int>::iterator it;
	for(it = fac.begin(); it != fac.end(); ++it)
	{
		int now = 1;
		while(it->second--)
		{
			now *= it->first;
		}
		multiply(now);
	}
	for(i = len-1; i >= 0;--i)
	{
		if(i==len-1)
		{
			printf("%d",num[i]);
		}
		else
		{
			printf("%06d",num[i]);
		}
	}
	puts("");
	return 0;
}

F

啥都不说，胡佬厉害！打印图形这种足以证明编程水平：

点击以显示 ⇲

点击以隐藏 ⇱ 

#include<stdio.h>
#include<math.h>
 
int a[105];
int b[105];
int c[105];
 
int main()
{
    int n, max1 = 0;
    scanf("%d",&n);
    int i;
    for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
        max1 =(max1>a[i])?max1:a[i];
    }
    for(i=0;i<n;i++)
	{
        if(a[i] == max1)
		{
			b[i] = 1;
		}
        c[i] = ceil(50.0 * a[i] / max1);
    }
    for(i=0;i<n;i++)
	{
        printf("+");
        int j;
        for(j=0;j<c[i];j++)
		{
			printf("-");
		}
        printf("+\n");
        printf("|");
        for(j=0;j<c[i]-1;j++)
		{
			printf(" ");
		}
        if(b[i] == 1)
		{
			printf("*|%d\n", a[i]);
		}
        else
		{
	        (c[i])?printf(" |%d\n", a[i]):printf("|%d\n", a[i]);
        }
        printf("+");
        for(j=0;j<c[i];j++)
		{
			printf("-");
		}
        printf("+\n");
    }
}

I

水。 

#include<stdio.h>
 
int main()
{
    printf("0.666667");
    return 0;
}

B

以下是补题。

看了标程。DFS的思路和我一模一样，但是后面最小生成树没有用Prim，而是选择了类似于Kruskal或者其他的优化方法。

点击以显示 ⇲

点击以隐藏 ⇱ 

#include<stdio.h>
 
#include<algorithm>
#include<vector>
 
using namespace std;
 
int fd(int N, int* A, int M, int* B, int K)
{
	if(N==0||M==0)
	{
		return 1<<K;
	}
	if(K==0)
	{
		return 0;
	}
	int X=lower_bound(A, A+N, ((A[0]>>K-1)|1)<<K-1)-A;
	int Y=lower_bound(B, B+M, ((B[0]>>K-1)|1)<<K-1)-B;
	if(X==0 && Y==M || X==N && Y==0)
	{
		return fd(N, A, M, B, K-1) + (1<<(K-1));
	}
	return min(fd(X, A, Y, B, K-1), fd(N-X, A+X, M-Y, B+Y, K-1));
}
 
long long solve(int N, int* A, int K)
{
	if(N == 1)
	{
		return 0;
	}
	int X =lower_bound(A, A+N, ((A[0] >> (K-1))|1) << (K-1)) - A;
	if(X==0 || X==N)
	{
		return solve(N, A, K-1);
	}
	return (1ll << (K-1)) + fd(X, A, N-X, A+X, K-1) + solve(X, A, K-1)+solve(N-X, A+X, K-1);
}
 
vector<pair<int,int> > G[202020];
 
int N, A[202020];
 
bool vis[202020];
 
void dfs(int x)
{
	vis[x] = 1;
	vector<pair<int,int> >::iterator e;
	for(e=G[x].begin();e!=G[x].end();e++)
	{
		if(!vis[e->first])
		{
			A[e->first] = A[x] ^ e->second;
			dfs(e->first);
		}
	}
}
 
int main()
{
	scanf("%d", &N);
	int i;
	for(i = 1; i < N; i++)
	{
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		G[u].push_back(make_pair(v,w));
		G[v].push_back(make_pair(u,w));
	}
	A[0] = 0;
	dfs(0);
	sort(A, A+N);
	N =unique(A, A+N) - A;
	printf("%lld\n", solve(N, A, 30));
	return 0;
}

C

比赛的时候没有看这道题，后来看了题解，才发现这是一道难得的生成函数的题目。

（好题啊）

点击以显示 ⇲

点击以隐藏 ⇱ 

#include<stdio.h>
 
#define mod 998244353
 
int fac[2020202],ffac[2020202];
 
int C(int n,int m)
{
	return 1ll*fac[n]*ffac[m]%mod*ffac[n-m]%mod;
}
 
int qpow(int n,int k)
{
	int ret = 1;
	while(k)
	{
		if(k&1)
		{
			ret = 1ll * ret * n % mod;
		}
		n = 1ll * n * n % mod;
		k /= 2;
	}
	return ret;
}
 
int main()
{
	fac[0] = 1;
	int i; 
	for(i = 1; i <= 2e6; i++)
	{
		fac[i] = 1ll * fac[i-1] * i % mod;
	}
	ffac[2000000] = qpow(fac[2000000], mod - 2);
	for(i = 2e6; i >= 1; i--)
	{
		ffac[i - 1] = 1ll * ffac[i] * i % mod;
	}
	int T;
	scanf("%d", &T);
	while(T--)
	{
		int N, M, K;
		scanf("%d%d%d", &N, &M, &K);
		int ans = 0;
		for(i = 0; i <=(N<M?N:M) - K; i++)
		{
			ans += 1ll * C(i + K - 1, K - 1) * C(N - i - 1, K - 1) % mod * C(M - i - 1, K - 1) % mod;
			if(ans >= mod)
			{
				ans -= mod;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}

J

有趣的计算几何。简单改了改标程，接下来基本可以简化成为C代码了。

点击以显示 ⇲

点击以隐藏 ⇱ 

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
 
#define pi acos((long double)-1)
#define eps 1e-11
 
long double aabs(long double x)
{
	return (long double)(x>0?x:(-x));
}
 
struct dn
{
	long double x,y;
};
 
struct dn operator+(struct dn u,struct dn v)
{
	return(struct dn){u.x+v.x,u.y+v.y};
}
 
struct dn operator-(struct dn u,struct dn v)
{
	return(struct dn){u.x-v.x,u.y-v.y};
}
 
struct dn operator*(long double u,struct dn v)
{
	return(struct dn){u*v.x,u*v.y};
}
 
struct dn operator/(struct dn u,long double v)
{
	return(struct dn){u.x/v,u.y/v};
}
 
long double operator/(struct dn u,struct dn v)
{
	return u.x*v.y-v.x*u.y;
}
 
long double operator*(struct dn u,struct dn v)
{
	return u.x*v.x+u.y*v.y;
}
 
long double dis(struct dn o)
{
	long double temp=o.x*o.x+o.y*o.y>0?o.x*o.x+o.y*o.y:0;
	return sqrt(temp);
}
 
struct dn dd(long double o)
{
	return(struct dn){cos(o),sin(o)};
}
 
struct dn g[5];
int top;
 
void swap(struct dn *a,struct dn *b)
{
	struct dn temp=(*a);
	(*a)=(*b);
	(*b)=temp;
}
 
void sswap(long double *a,long double *b)
{
	long double temp=(*a);
	(*a)=(*b);
	(*b)=temp;
}
 
int jd(struct dn o1,long double r1,struct dn o2,long double r2,struct dn p,struct dn q,struct dn *u,struct dn *v)
{
	top=0;
	struct dn o=(o1-p)*(q-p)*(q-p)/((q-p)*(q-p))+p;
	if(dis(o-o1)<r1-eps)
	{
		g[top]=(o+sqrt(r1*r1-(o1-o)*(o1-o))*(q-p)/dis(q-p));
		top++;
		g[top]=(o-sqrt(r1*r1-(o1-o)*(o1-o))*(q-p)/dis(q-p));
		top++;
	}
	o=(o2-p)*(q-p)*(q-p)/((q-p)*(q-p))+p;
	if(dis(o-o2)<r2-eps)
	{
		g[top]=(o+sqrt(r2*r2-(o2-o)*(o2-o))*(q-p)/dis(q-p));
		top++;
		g[top]=(o-sqrt(r2*r2-(o2-o)*(o2-o))*(q-p)/dis(q-p));
		top++;
	}
	int i; 
	for(i=0;i<top;i++)
	{
		int ok=0;
		int j;
		for(j=0;j<i;j++)
		{
			if(dis(g[i]-g[j])<eps)
			{
				ok=1;
			}
		}
		if(dis(g[i]-o1)>r1+eps||dis(g[i]-o2)>r2+eps)
		{
			ok=1;
		}
		if((g[i]-p)*(q-p)<-eps)
		{
			ok=1;
		}
		if(ok)
		{
			swap(&g[i],&g[top-1]);
			top--;
			i--;
		}
	}
	if(top>0)
	{
		*u=g[0];
	}
	if(top>1)
	{
		*v=g[1];
	}
	if(top==2&&((*u)-p)*(q-p)>((*v)-p)*(q-p))
	{
		swap(&(*u),&(*v));
	}
	return top;
}
 
long double arc(struct dn o,long double r,struct dn u,struct dn v)
{
	long double w=(v-o)*(u-o)/r/r;
	w=(long double)(w>(-1)?w:(-1));
	w=(long double)(w<1?w:1);
	if((o-u)/(v-u)>0)
	{
		return 0.5*acos(w)*r*r-0.5*aabs((v-o)/(u-o));
	}
	else
	{
		return pi*r*r-0.5*acos(w)*r*r+0.5*aabs((v-o)/(u-o));
	}
}
 
long double ar(struct dn o1,long double r1,struct dn o2,long double r2,struct dn u,struct dn v,int www)
{
	if(dis(o2-o1)<r2-r1+eps)
	{
		if(www)
		{
			return pi*r1*r1;
		}
		else
		{
			return arc(o1,r1,u,v);
		}
	}
	struct dn d1,d2;
	struct dn o=(r1*r1+(o2-o1)*(o2-o1)-r2*r2)/(2*r1*dis(o2-o1))*r1*(o2-o1)/dis(o2-o1);
	struct dn w=sqrt(r1*r1-o*o)*o/dis(o);
	d1=o1+o+(struct dn){w.y,-w.x};
	d2=o1+o-(struct dn){w.y,-w.x};
	if(o*(o2-o1)<eps)
	{
		swap(&d1,&d2);
	}
	long double aa=arc(o1,r1,d2,d1)+arc(o2,r2,d1,d2);
	if(www)
	{
		return aa;
	}
	if((u-d1)/(d2-d1)<eps&&(v-d1)/(d2-d1)<eps)
	{
		if((0.5*(d1+d2)-u)/(v-u)>eps||(0.5*(d1+d2)-u)/(v-u)>-eps&&(v-u)*(d2-d1)>0)
		{
			return arc(o2,r2,u,v);
		}
		else
		{
			return aa-arc(o2,r2,v,u);
		}
	}
	if((u-d1)/(d2-d1)>-eps&&(v-d1)/(d2-d1)>-eps)
	{
		if((0.5*(d1+d2)-u)/(v-u)>eps||(0.5*(d1+d2)-u)/(v-u)>-eps&&(v-u)*(d1-d2)>0)
		{
			return arc(o1,r1,u,v);
		}
		else
		{
			return aa-arc(o1,r1,v,u);
		}
	}
	if((u-d1)/(d2-d1)>-eps)
	{
		return arc(o1,r1,u,d1)+arc(o2,r2,d1,v)+0.5*aabs((u-d1)/(v-d1));
	}
	else
	{
		return arc(o1,r1,d2,v)+arc(o2,r2,u,d2)+0.5*aabs((u-d2)/(v-d2));
	}
}
 
long double cc(long double p,long double q,struct dn o1,long double r1,struct dn o2,long double r2)
{
	if(dis(o1-o2)>r1+r2-eps)
	{
		return 0;
	}
	if(r1>r2)
	{
		swap(&o1,&o2);
		sswap(&r1,&r2);
	}
	if(r1<eps)
	{
		return 0;
	}
	if(dis(o1)<r1-eps&&dis(o2)<r2-eps)
	{
		struct dn u,v;
		jd(o1,r1,o2,r2,(struct dn){0,0},dd(p),&u,&v);
		jd(o1,r1,o2,r2,(struct dn){0,0},dd(q),&v,&u);
		return dis(u-v)<eps?0:ar(o1,r1,o2,r2,v,u,0)+0.5*aabs(u/v);
	}
	else
	{
		struct dn u1,u2,v1,v2;
		int ok1,ok2;
		ok1=jd(o1,r1,o2,r2,(struct dn){0,0},dd(p),&u1,&u2);
		ok2=jd(o1,r1,o2,r2,(struct dn){0,0},dd(q),&v1,&v2);
		if(ok1==1)
		{
			ok1=0;
		}
		if(ok2==1)
		{
			ok2=0;
		}
		long double aa=ar(o1,r1,o2,r2,(struct dn){0,0},(struct dn){0,0},1);
		if(!ok1&&!ok2)
		{
			struct dn o;
			if(dis(o1-o2)<r2-r1+eps)
			{
				o=o1;
			}
			else
			{
				o=(r1*r1+(o2-o1)*(o2-o1)-r2*r2)/(2*r1*dis(o2-o1))*r1*(o2-o1)/dis(o2-o1)+o1;
			}
			if(o/dd(p)<-eps&&o/dd(q)>eps)
			{
				return aa;
			}
			else
			{
				return 0;
			}
		}
		if(ok1)
		{
			aa-=ar(o1,r1,o2,r2,u2,u1,0);
		}
		if(ok2)
		{
			aa-=ar(o1,r1,o2,r2,v1,v2,0);
		}
		return aa;
	}
}
 
int main()
{
	int t=100000;
	scanf("%d",&t);
	int tt;
	for(tt=1;tt<=t;tt++)
	{
		double a,a1,a2,z1,z2,r1,r2;
		a=rand()%89+1;
		a1=rand()%360;
		a2=rand()%360;
		z1=rand()%2001;
		z2=rand()%2001;
		r1=rand()%2001;
		r2=rand()%2001;
		scanf("%lf%lf%lf%lf%lf%lf%lf",&a,&a1,&a2,&z1,&z2,&r1,&r2);
		a2=a2-a1+360;
		if(a2>=360)
		{
			a2-=360;
		}
		a1=0;
		if(a2>180)
		{
			a2=360-a2;
		}
		a=sin(a/180*pi);
		long double ans=0;
		ans+=cc(0,a2/180*pi*a,z2*dd(a2/180*pi*a),r2,z1*dd(0),r1);
		ans+=cc(0,(180-a2)/180*pi*a,z2*dd(0),r2,z1*dd(-a2/180*pi*a),r1);
		ans+=cc(0,a2/180*pi*a,z2*dd(-(180-a2)/180*pi*a),r2,z1*dd(pi*a),r1);
		ans+=cc(0,(180-a2)/180*pi*a,z2*dd(pi*a),r2,z1*dd((180-a2)/180*pi*a),r1);
		printf("%.30f\n",(double)ans);
	}
}
2020-2021/teams/namespace/牛客多校第五场.1595928061.txt.gz · 最后更改: 2020/07/28 17:21 由 great_designer

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