2017 ACM/ICPC Asia Regional Qingdao Online

時間  2019-12-02
標籤 acm icpc asia regional qingdao online 欄目 應用數學 简体版
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Apple

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 982    Accepted Submission(s): 323

Problem Description
Apple is Taotao's favourite fruit. In his backyard, there are three apple trees with coordinates  (x1,y1), (x2,y2), and (x3,y3). Now Taotao is planning to plant a new one, but he is not willing to take these trees too close. He believes that the new apple tree should be outside the circle which the three apple trees that already exist is on. Taotao picked a potential position (x,y) of the new tree. Could you tell him if it is outside the circle or not?
 

Input
The first line contains an integer  T, indicating that there are T(T≤30) cases.
In the first line of each case, there are eight integers x1,y1,x2,y2,x3,y3,x,y, as described above.
The absolute values of integers in input are less than or equal to 1,000,000,000,000.
It is guaranteed that, any three of the four positions do not lie on a straight line.
 

Output
For each case, output "Accepted" if the position is outside the circle, or "Rejected" if the position is on or inside the circle.
 

Sample Input
3 -2 0 0 -2 2 0 2 -2 -2 0 0 -2 2 0 0 2 -2 0 0 -2 2 0 1 1
 

Sample Output
Accepted Rejected Rejected
 

Source
2017 ACM/ICPC Asia Regional Qingdao Online
 
 

圓的反演，java大數走一波php

其實用外接圓半徑及點那個不復雜的東西也能夠玩java

import java.io.*;
import java.util.*;
import java.math.*;

public class Main
{
    static public BigInteger xu[]=new BigInteger[4];
    static public BigInteger xd[]=new BigInteger[4];
    static public BigInteger yu[]=new BigInteger[4];
    static public BigInteger yd[]=new BigInteger[4];
    static public void main(String[] args)
    {
        Scanner cin=new Scanner(System.in);
        int T=cin.nextInt();
        while(T-->0)
        {
            for(int i=0;i<4;i++)
            {
                xu[i]=cin.nextBigInteger();
                yu[i]=cin.nextBigInteger();
                xd[i]=BigInteger.ONE;
                yd[i]=BigInteger.ONE;
            }
            for(int i=1;i<4;i++)
            {
                xu[i]=xu[i].subtract(xu[0]);
                yu[i]=yu[i].subtract(yu[0]);
            }
            for(int i=1;i<4;i++)
            {
                xd[i]=xu[i].multiply(xu[i]).add(yu[i].multiply(yu[i]));
                yd[i]=xu[i].multiply(xu[i]).add(yu[i].multiply(yu[i]));
            }
            BigInteger t=xu[1].multiply(yu[2]).subtract(yd[1].multiply(xd[2]));
            t=t.subtract(yu[1].multiply(xu[2]).subtract(xd[1].multiply(yd[2])));
            if(t.compareTo(BigInteger.ZERO)<0)
            {
                t=xu[1];xu[1]=xu[2];xu[2]=t;
                t=yu[1];yu[1]=yu[2];yu[2]=t;
                t=xd[1];xd[1]=xd[2];xd[2]=t;
                t=yd[1];yd[1]=yd[2];yd[2]=t;
            }
            for(int i=2;i<=3;i++)
            {
                xu[i]=xu[i].multiply(xd[1]).subtract(xu[1].multiply(xd[i]));
                xd[i]=xd[i].multiply(xd[1]);
                yu[i]=yu[i].multiply(yd[1]).subtract(yu[1].multiply(yd[i]));
                yd[i]=yd[i].multiply(yd[1]);
            }
            t=xu[3].multiply(yu[2]).subtract(yd[3].multiply(xd[2]));
            t=t.subtract(yu[3].multiply(xu[2]).subtract(xd[3].multiply(yd[2])));
            if(t.compareTo(BigInteger.ZERO)>=0)System.out.println("Rejected");
            else System.out.println("Accepted");
        }
    }
}

 

The Dominator of Strings

Time Limit: 3000/3000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 6778    Accepted Submission(s): 713


node

Problem Description
Here you have a set of strings. A dominator is a string of the set dominating all strings else. The string  S is dominated by T if S is a substring of T.
 

 

Input
The input contains several test cases and the first line provides the total number of cases.
For each test case, the first line contains an integer  N indicating the size of the set.
Each of the following N lines describes a string of the set in lowercase.
The total length of strings in each case has the limit of 100000.
The limit is 30MB for the input file.
 

 

Output
For each test case, output a dominator if exist, or No if not.
 

 

Sample Input
3
10
you
better
worse
richer
poorer
sickness
health
death
faithfulness
youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
5
abc
cde
abcde
abcde
bcde
3
aaaaa
aaaab
aaaac
 

 

Sample Output
youbemyweddedwifebetterworsericherpoorersicknesshealthtilldeathdouspartandpledgeyoumyfaithfulness
abcde
No
從較長串中找子串的
自動機的複雜度過高，挑戰上的代碼就能夠過得
這個暴力strstr也能夠過，也就是KMP，必定是我太醜了 
#include<bits/stdc++.h>
  
using namespace std;  
const int MAXN = 100010;  
int n,k;  
int Paiming[MAXN+1],tmp[MAXN+1];  
int flag;  
bool comp_sa(int i, int j)  
{  
    if(Paiming[i] != Paiming[j])   
        return Paiming[i] < Paiming[j];  
    else{
        int ri = i+k <= n? Paiming[i+k] : -1;  
        int rj = j+k <= n? Paiming[j+k] : -1;  
        return ri < rj;     
    }  
}  
  
void calc_sa(string &S, int *sa) 
{  
    n = S.size();  
    for(int i = 0; i <= n; i++)  
    {  
        sa[i] = i;  
        Paiming[i] = i < n ? S[i] : -1;  
    }  
      
    for( k =1; k <= n; k *= 2)  
    {  
        sort(sa,sa+n+1,comp_sa);  
        tmp[sa[0]] = 0;  
        for(int i = 1; i <= n; i++)  
        {  
            tmp[sa[i]] = tmp[sa[i-1]] + (comp_sa(sa[i-1],sa[i]) ? 1: 0);  
        }  
        for(int i = 0; i <= n; i++)  
        {  
            Paiming[i] = tmp[i];  
        }  
    }  
}  
  
int SuffixArrayMatch(string &S, int *sa, string T)  
{  
    
    int lhs = 0, rhs = S.size();  
    while(rhs - lhs > 1)  
    {  
        int mid = (lhs + rhs)>>1;  
        if(S.compare(sa[mid],T.size(),T) < 0) lhs = mid;
        else rhs=mid;   
    }
    return S.compare(sa[rhs],T.size(),T) == 0;  
}  
int main()  
{  
    int t;
    ios::sync_with_stdio(false);
    cin>>t;        
    string s[100010],longs;    
    while(t--){
        int n,l=-1,p=-1,i; 
        cin>>n;
        memset(Paiming,0,sizeof Paiming);
        memset(tmp,0,sizeof tmp);
        for(i=0;i<n;i++){
            cin>>s[i];
            int len=s[i].size();
            if(len>l){
                l=len;
                longs=s[i];
                p=i;
            }
        }
        if(n==1){
            cout<<longs<<endl;
            continue;
        }   
        int *sa = new int[longs.size()+1];  
        calc_sa(longs,sa);
        for(i=0;i<n;i++){
            if(p==i)continue;    
            if(!SuffixArrayMatch(longs,sa,s[i]))
            break;  
        } 
        //delete [] sa;  
        sa = NULL; 
        if(i>=n){
            cout<<longs<<endl;
        }else {
            cout<<"No"<<endl;
        }      
    }   
}

 

Chinese Zodiac

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2451    Accepted Submission(s): 1645

Problem Description
The Chinese Zodiac, known as Sheng Xiao, is based on a twelve-year cycle, each year in the cycle related to an animal sign. These signs are the rat, ox, tiger, rabbit, dragon, snake, horse, sheep, monkey, rooster, dog and pig.
Victoria is married to a younger man, but no one knows the real age difference between the couple. The good news is that she told us their Chinese Zodiac signs. Their years of birth in luner calendar is not the same. Here we can guess a very rough estimate of the minimum age difference between them.
If, for instance, the signs of Victoria and her husband are ox and rabbit respectively, the estimate should be  2 years. But if the signs of the couple is the same, the answer should be 12 years.
 

Input
The first line of input contains an integer  T (1≤T≤1000) indicating the number of test cases.
For each test case a line of two strings describes the signs of Victoria and her husband.
 

Output
For each test case output an integer in a line.
 

Sample Input
3
ox rooster
rooster ox
dragon dragon
 

Sample Output
8
4
12
中國生肖，隨手模擬就能夠了 
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
using namespace std;
int main()
{
    char m[12][10]={
        "rat", "ox", "tiger", "rabbit", "dragon", "snake", "horse", "sheep", "monkey", "rooster", "dog" , "pig"
    };
    int t,s;
    scanf("%d",&t);
    while(t--){
        char p[10],q[10];
        scanf("%s%s",&p,&q);
        if(strcmp(p,q)==0){
            puts("12");
        }else {
            int i,pp,qq;
            for(i=0;i<12;i++){
                if(strcmp(m[i],p)==0){
                    pp=i;
                }
                if(strcmp(m[i],q)==0){
                    qq=i;
                }
            }
            s=(pp-qq);
            if(s>0){
                s=s-12;
            }
            printf("%d\n",-s);
        }
    }
}

 

Smallest Minimum Cut

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3297    Accepted Submission(s): 602

Problem Description
Consider a network  G=(V,E) with source s and sink t. An s-t cut is a partition of nodes set V into two parts such that s and t belong to different parts. The cut set is the subset of E with all edges connecting nodes in different parts. A minimum cut is the one whose cut set has the minimum summation of capacities. The size of a cut is the number of edges in the cut set. Please calculate the smallest size of all minimum cuts.
 

Input
The input contains several test cases and the first line is the total number of cases  T (1≤T≤300).
Each case describes a network G, and the first line contains two integers n (2≤n≤200) and m (0≤m≤1000) indicating the sizes of nodes and edges. All nodes in the network are labelled from 1 to n.
The second line contains two different integers s and t (1≤s,t≤n) corresponding to the source and sink.
Each of the next m lines contains three integers u,v and w (1≤w≤255) describing a directed edge from node u to v with capacity w.
 

Output
For each test case, output the smallest size of all minimum cuts in a line.
 

Sample Input
2
4 5
1 4
1 2 3
1 3 1
2 3 1
2 4 1
3 4 2
4 5
1 4
1 2 3
1 3 1
2 3 1
2 4 1
3 4 3
Sample Output
2
3
求點s到點t的最小割，這個網上就有一道同樣的題吧
dinic是過不去的，要用sap算法 
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAXN 2333
#define MAXM 2333333
struct Edge
{
    int v,next;
    ll cap;
} edge[MAXM];
int head[MAXN],pre[MAXN],cur[MAXN],level[MAXN],gap[MAXN],NV,NE,n,m,vs,vt;
void ADD(int u,int v,ll cap,ll cc=0)
{
    edge[NE].v=v;
    edge[NE].cap=cap;
    edge[NE].next=head[u];
    head[u]=NE++;

    edge[NE].v=u;
    edge[NE].cap=cc;
    edge[NE].next=head[v];
    head[v]=NE++;
}
ll SAP(int vs,int vt)
{
    memset(pre,-1,sizeof(pre));
    memset(level,0,sizeof(level));
    memset(gap,0,sizeof(gap));
    for(int i=0; i<=NV; i++)cur[i]=head[i];
    int u=pre[vs]=vs;
    ll aug=-1,maxflow=0;
    gap[0]=NV;
    while(level[vs]<NV)
    {
loop:
        for(int &i=cur[u]; i!=-1; i=edge[i].next)
        {
            int v=edge[i].v;
            if(edge[i].cap&&level[u]==level[v]+1)
            {
                aug==-1?aug=edge[i].cap:aug=min(aug,edge[i].cap);
                pre[v]=u;
                u=v;
                if(v==vt)
                {
                    maxflow+=aug;
                    for(u=pre[u]; v!=vs; v=u,u=pre[u])
                    {
                        edge[cur[u]].cap-=aug;
                        edge[cur[u]^1].cap+=aug;
                    }
                    aug=-1;
                }
                goto loop;
            }
        }
        int minlevel=NV;
        for(int i=head[u]; i!=-1; i=edge[i].next)
        {
            int v=edge[i].v;
            if(edge[i].cap&&minlevel>level[v])
            {
                cur[u]=i;
                minlevel=level[v];
            }
        }
        if(--gap[level[u]]==0)break;
        level[u]=minlevel+1;
        gap[level[u]]++;
        u=pre[u];
    }
    return maxflow;
}

int main()
{
    int T,u,v,w;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        scanf("%d%d",&vs,&vt);
        NV=n,NE=0;
        memset(head,-1,sizeof(head));
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            ADD(u,v,(ll)w*MAXM+1);
        }
        ll ans=SAP(vs,vt);
        printf("%d\n",ans%MAXM);
    }
    return 0;
}

 

A Cubic number and A Cubic Number

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4947    Accepted Submission(s): 1346

Problem Description
A cubic number is the result of using a whole number in a multiplication three times. For example,  3×3×3=27 so 27 is a cubic number. The first few cubic numbers are 1,8,27,64 and 125. Given an prime number p. Check that if p is a difference of two cubic numbers.
 

Input
The first of input contains an integer  T (1≤T≤100) which is the total number of test cases.
For each test case, a line contains a prime number p (2≤p≤1012).
 

Output
For each test case, output 'YES' if given  p is a difference of two cubic numbers, or 'NO' if not.
 

Sample Input
10
2
3
5
7
11
13
17
19
23
29
 

Sample Output
NO
NO
NO
YES
NO
NO
NO
YES
NO
這個我是機房第一個作出來的啊，讓你看一個數是否是兩個數的立方差
可是這個數是質數，因式分解判斷另外一段就好的
直接二分答案存不存在就好 
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
bool la(LL x)
{
    LL l=1,r=1e6+5;
    while(l<=r)
    {
        LL mi=(l+r)/2;
        LL y=mi-1;
        if(mi*mi+y*y+mi*y==x)
            return 1;
        else if(mi*mi+y*y+mi*y<x)
            l=mi+1;
        else r=mi-1;
    }
    return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        LL n;
        scanf("%lld",&n);
        printf("%s\n",la(n)?"YES":"NO");
    }
    return 0;
}
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