2017青岛区域赛（部分题解）

vjudge上题目链接：VJ

//垃圾
#include<bits/stdc++.h>
using namespace std;
int judge(char s[])
{
	if(strcmp(s,"rat")==0)
	   return 1;
	else if(strcmp(s,"ox")==0)
	   return 2;
	else if(strcmp(s,"tiger")==0)
	   return 3;
	else if(strcmp(s,"rabbit")==0)
	   return 4;
	else if(strcmp(s,"dragon")==0)
	   return 5;
	else if(strcmp(s,"snake")==0)
	   return 6;
	else if(strcmp(s,"horse")==0)
	   return 7;
	else if(strcmp(s,"sheep")==0)
	   return 8;
	else if(strcmp(s,"monkey")==0)
	   return 9;
	else if(strcmp(s,"rooster")==0)
	   return 10;
	else if(strcmp(s,"dog")==0)
	   return 11;
	if(strcmp(s,"pig")==0)
	   return 12;
}
int main()
{
	int t;
	char s1[10];
	char s2[10];
	scanf("%d",&t);
	while(t--)
	{
		cin>>s1>>s2;
		if(strcmp(s1,s2)==0)
		   {
		   	  printf("12\n");
		   	  continue;
		   }
		else
		{
			int ans1=judge(s1);
			int ans2=judge(s2);
			if(ans2>ans1)
			   cout<<ans2-ans1<<endl;
			else
			   cout<<ans2-ans1+12<<endl;
		}
	}
	return 0;
}

还是用map好，做题都蒙圈了。。。

#include<bits/stdc++.h>
using namespace std;
string a[15]={"rat","ox","tiger","rabbit","dragon","snake","horse","sheep","monkey","rooster","dog","pig"};
map<string,int> mmp;
void init()
{
	
	for(int i=1;i<=12;i++)
	{
		mmp[a[i]]=i;
	}
	
}
int main()
{
	int T;
	init();
	scanf("%d",&T);
	while(T--)
	{
		string s1,s2;
		cin >> s1 >> s2;
		int x1=mmp[s1],x2=mmp[s2];
		if(x1 == x2)
		{
			printf("12\n");
		}
		else if(x2 > x1)
			printf("%d\n",x2-x1);
		else
			printf("%d\n",(x2+12-x1)%12);
	}
	
	
	return 0;
}

K A Cubic number and A Cubic Number

主要还是思维，首先知道公式：a^3-b^3=（a-b）*(a^2+ab+b^2)，p是素数，所以（a-b）=1很重要的点，知道这个问题就解决了。

#include<bits/stdc++.h>
using namespace std;
#define ll long long
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
	{
        ll n;
        scanf("%lld",&n);
        ll  i,flag=0,t=1;
        while(3*t*t+3*t+1<=n)
		{
            if(3*t*t+3*t+1==n)
			{
                printf("YES\n");
                flag=1;
            }
            t++;
        }
        if(!flag)
            printf("NO\n");
 
    }
    return 0;
}

A Apple

高进度用java，看到表达式，开始怀疑人生。

import java.math.*;
import java.util.*;
//a=((y2-y1)*(y3*y3-y1*y1+x3*x3-x1*x1)-(y3-y1)*(y2*y2-y1*y1+x2*x2-x1*x1))/(2.0*((x3-x1)*(y2-y1)-(x2-x1)*(y3-y1)));
//b=((x2-x1)*(x3*x3-x1*x1+y3*y3-y1*y1)-(x3-x1)*(x2*x2-x1*x1+y2*y2-y1*y1))/(2.0*((y3-y1)*(x2-x1)-(y2-y1)*(x3-x1)));
public class Main {
    public static void main(String[] args) {
    	 BigDecimal x1,x2,x3,y1,y2,y3,x,y,x0,y0,r;
         BigDecimal a,b,c,d,e,f,g,num,ans;
         Scanner cin = new Scanner(System.in);
         int n;
         n=cin.nextInt();
         while(n>0)
         {
             x1  = cin.nextBigDecimal();
             y1  = cin.nextBigDecimal();
             x2  = cin.nextBigDecimal();
             y2  = cin.nextBigDecimal();
             x3  = cin.nextBigDecimal();
             y3  = cin.nextBigDecimal();
             x  = cin.nextBigDecimal();
             y  = cin.nextBigDecimal();
             a = x3.subtract(x2).multiply(BigDecimal.valueOf(2));// 2*(x3-x2)
             b = y3.subtract(y2).multiply(BigDecimal.valueOf(2));// 2*(y3-y2)
             c = x3.pow(2).subtract(x2.pow(2)).add(y3.pow(2).subtract(y2.pow(2)));
               //x3^2-x2^2+(y3^2-y2^2)
             e = x2.subtract(x1).multiply(BigDecimal.valueOf(2));//2*(x2-x1)
             f = y2.subtract(y1).multiply(BigDecimal.valueOf(2));//2*(y2-y1)
             g = x2.pow(2).subtract(x1.pow(2)).add(y2.pow(2).subtract(y1.pow(2)));//x2^2-(x1^2)+(y2^2-y1^2)
             x0=g.multiply(b).subtract(c.multiply(f)).divide(e.multiply(b).subtract(a.multiply(f)));//(g*b-(c*f))/(e*b-a*f)
             y0=a.multiply(g).subtract(c.multiply(e)).divide(a.multiply(f).subtract(b.multiply(e)));//(a*g-c*e)/(a*f-b*e)
             num = (x1.subtract(x0)).pow(2).add((y1.subtract(y0)).pow(2));//(x1-x0)^2+(y1-y0)^2
             ans = (x.subtract(x0)).pow(2).add((y.subtract(y0)).pow(2));
             if (ans.compareTo(num)>0) {
                 System.out.println("Accepted");
             }
             else
                 System.out.println("Rejected");
            n--;
         }
    }

}

C The Dominator of Strings

题意还是比较好明白的，就是看看最长的字符串中是不是含有其它字符串。可是普通的办法肯定会超时，不容易实现，再一次见证了JAVA的强大。

这有BufferedReader 的介绍：BufferedReader

借鉴

import java.io.*;
public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
        String str[]=new String[100010];
        String t=bf.readLine();
        int T=Integer.parseInt(t);
        while(T>0)
        {
        	String n=bf.readLine();
        	int N=Integer.parseInt(n);
    		String news="";
        	for(int i=0;i<N;i++)
        	{
        		str[i]=bf.readLine();
        		if(str[i].length()>news.length())
        			news=str[i];
        	}
        	int flag=1;
        	for(int i=0;i<N;i++)
        	{
        		if(!news.contains(str[i]))
        		{
        			flag=0;
        			break;
        		}
        	}
        	if(flag==1) System.out.println(news);
        	else System.out.println("No");
        	T--;
        }
        
    }
}

find()函数也是以绝，学习find()点击转到。

#include<bits/stdc++.h>
using namespace std;
string str[100010];
int t,n;
int main()
{
	ios::sync_with_stdio(false);
	int n,t;
    scanf("%d",&t);
    while(t--)
    {
        int flag=1;
        scanf("%d",&n);
        getchar();
        for(int i=1;i<=n;i++)
        {
            cin>>str[i];
        }
        for(int i=1;i<=n;i++)
        {
            int ans=0;
            for(int j=1;j<=n;j++)
            {
                if(str[i].find(str[j])!=-1) 
				  ans++;
                else break;
            }
            if(ans==n)
            {
                cout<<str[i]<<endl;
                flag=0;
                break;
            }
        }
        if(flag) printf("No\n");
    }
    return 0;
}

2017青岛区域赛 （部分题解）

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2017青岛区域赛（部分题解）