武汉工程大学2020GPLT选拔赛（重现赛） 部分题解
A:L1-1 I LOVE WIT
很简单的一个模拟题
只需要按照他的条件输出就好了，记得那两行空白是有空格的哦。
import java.math.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;
public class Main {
public static void main(String args[])throws IOException
{
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
out.println( "I");
out.println(" ");
out.println(" L");
out.println(" O");
out.println(" V");
out.println(" E");
out.println(" ");
out.println(" W");
out.println(" I");
out.print( " T");
out.flush();
}
} B:L1-2 单位换算 只需要将n乘以304.8即可，注意要分别是否小数点后是否为0，可以用 (n*304.8)==(int)(n*304.8)进行判断，然后用format取小数点后一位即可。
import java.math.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;
public class Main {
public static void main(String args[])throws IOException
{
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
in.nextToken();
double n = in.nval;
if((n*304.8)==(int)(n*304.8))
out.println((int)(n*304.8));
else{
out.println(String.format("%.1f",n*304.8));
}
out.flush();
} }
C:L1-3 Pokémon 这道题有一点陷阱，就是闪光的概率为指定v型的概率乘以0.01，要是普通的概率为指定v型概率乘以0.99，而不是1.
如果仔细看到这里的话就不会出错啦。
D:L1-4 颠倒阴阳
模拟题，先把n每一位遍历一遍，如果是0就换成1，相反。
然后再把s字符串补0补齐32位。
然后遍历&1==1的话就加到sum中。
记得开long
import java.math.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;
public class Main {
public static void main(String args[])throws IOException
{
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
in.nextToken();
long n = (long)in.nval;
String s="";
while(n>0)
{
if((n&1)==1)
s = s+0;
else{
s = s+1;
}
n>>=1;
}
for(int i=s.length();i<32;i++)
{
s+=0;
}
long sum=0,x=1;
for(int i=31;i>=0;i--)
{
if(s.charAt(i)=='1')
sum+=(x);
x*=2;
}
out.print(sum);
out.flush();
} }
E:L1-5 演唱会 直接把时间写成一个六位数，分别于173727和193727比较即可；
import java.math.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.*;
public class Main {
public static void main(String args[])throws IOException
{
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
Scanner input = new Scanner(System.in);
String s = input.nextLine();
int p1=0,p2=0;
for(int i=0;i<s.length();i++)
{
if(p1!=0&&s.charAt(i)==':')
p2 = i;
if(p1==0&&s.charAt(i)==':')
p1 = i;
}
String xx = s.substring(0,p1),yy = s.substring(p1+1,p2),zz = s.substring(p2+1);
int x = Integer.parseInt(xx)*10000+Integer.parseInt(yy)*100+Integer.parseInt(zz);
if(x<173727)
System.out.print("arrive on time");
if(x>=173727&&x<193727)
System.out.print("arrive late");
if(x>=193727)
System.out.print("too late");
} }
F：L1-6 分鸽子 典型的二分模板题
和之前那个每日一题里的华华给月月礼物基本一样，分小木棍那个。
在1和最大的值之间二分查找，够分的肉的最大值输出即可。
import java.util.*;
import java.math.*;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.PrintWriter;
public class Main {
public static void main(String args[])throws IOException
{
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
in.nextToken();
int n = (int)in.nval;
in.nextToken();
int m = (int)in.nval;
int num[] = new int[n];
for(int t=0;t<n;t++)
{
in.nextToken();
num[t] = (int)in.nval;
}
int sum=0;
Arrays.sort(num);
int l=1,r=num[n-1],mid=(l+r)>>1,max=0;
while(l<=r)
{
mid=(l+r)>>1;
if(check(mid,num)>=m)
{
max = Math.max(max,mid);
l = mid+1;
}
else{
r = mid-1;
}
}
out.print(max);
out.flush();
}
public static int check(int p,int num[])
{
int sum=0;
for(int i=0;i<num.length;i++)
{
sum+=(num[i]/p);
}
return sum;
}
} G:L1-7 拼接梯子 分两种情况：
1.不能被2整除也就是最后一位二进制为1的，这样不可能凑成；
2.能被2整除的，直接看最大的那一位二进制的位置是否在拥有的种类中。
附加一个快速幂。
import java.util.*;
import java.math.*;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.io.OutputStreamWriter;
import java.io.BufferedReader;
import java.io.PrintWriter;
public class Main {
public static void main(String args[])throws IOException {
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
in.nextToken();
int k = (int) in.nval;
in.nextToken();
long l = (long) in.nval;
long sum = 0, L = l, p = 0;
if ((l & 1) == 1)
out.print("No");
else {
while (L > 0) {
if ((L & 1) == 1)
sum = p;
p++;
L >>= 1;
}
if (sum <= k) {
out.println("Yes");
out.println(pow(2, sum));
} else
out.print("No");
}
out.flush();
}
public static long pow(long n,long x)
{
long temp=n,sum=1;
while(x>0)
{
if((x&1)==1)
{
sum*=temp;
}
temp*=temp;
x>>=1;
}
return sum;
}
} 
