Group programming TIANTI competition exercise - continuously updating

1、L1-003 Single digit statistics fraction 15

Input format ：
Each input contains 1 Test cases , I.e. no more than one 1000 Bit positive integer N.

Output format ：
Yes N Each different digit in , With D:M Output the digit in one line D And in N Is the number of times M. Required press D Ascending output of .
Examples ：

100311

0:2
1:3
3:1

AC Code ：

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    
    char N[1005];
    scanf("%s",N);
    int num[10]={
    0};
    for(int i=0;i<10;i++)
    {
    
        for(unsigned j=0;j<strlen(N);j++)
        {
    
            if((N[j]-48)==i)// Convert characters to numbers 
            {
    
                num[i]++;
            }
        }
    }
    for(int i=0;i<10;i++)
    {
    
        if(num[i]!=0)
        {
    
            printf("%d:%d\n",i,num[i]);
        }
    }
    return 0;
}

1、L1-005 Test seat number fraction 15

Every PAT Candidates are assigned two seat numbers when they take the exam , One is a test seat , One is the exam seat . Under normal circumstances , Candidates will get the seat number of the test machine before entering , After entering the test state , The system will display the test seat number of the candidate , Candidates need to change their seats during the examination . But some candidates are late , The test run is over , They can only ask you for help with the number of the test seat they have received , Find out their test seat number from the background .

Input format ：
The first line of input gives a positive integer N（≤1000）, And then N That's ok , Each line gives a candidate's information ： Ticket number Test seat number Test seat number . Among them, the examination Permit No. is issued by 16 Digit composition , Seat from 1 To N Number . Input to make sure that everyone's admission number is different , And at no time will two people be assigned to the same seat .

After candidate information , Give a positive integer M（≤N）, In the next line M Test seat numbers to be inquired , Space off .

Output format ：
Corresponding to each seat number to be inquired , Output the candidate's pass number and seat number in one line , Intermediate use 1 Space separation .

Examples ：

4
3310120150912233 2 4
3310120150912119 4 1
3310120150912126 1 3
3310120150912002 3 2
2
3 4

3310120150912002 2
3310120150912119 1

AC Code ：

#include<iostream>
#include<cstring>
using namespace std;
// Use structures to store information 
struct Node
{
    
  char id[18];// Ticket number 
  int sj;// Test seat number 
  int ks;// Test seat number 
};
int main()
{
    
    int N;
    scanf("%d",&N);
    struct Node node[N];// Define an array of structures , Use struct
    char id[18];
    int sj,ks;
    // Enter and save data 
    for(int i=0;i<N;i++)
    {
    
        scanf("%s%d%d",id,&sj,&ks);
        strcpy(node[i].id,id);
        node[i].sj=sj;
        node[i].ks=ks;
    }
    int M;
    scanf("%d",&M);
    // Query sequentially 
    for(int i=0;i<M;i++)
    {
    
        int k=0;
        scanf("%d",&k);
        for(int j=0;j<N;j++)
        {
    
            if(node[j].sj==k)
            {
    
                printf("%s %d\n",node[j].id,node[j].ks);
            }
        }
    }
    return 0;
}

1、L1-007 Read numbers fraction 10

Enter an integer , Output the Pinyin corresponding to each number . When the integer is negative , First, the output fu word . The Pinyin corresponding to ten numbers is as follows ：

0: ling
1: yi
2: er
3: san
4: si
5: wu
6: liu
7: qi
8: ba
9: jiu

Input format ：
Input gives an integer on a line , Such as ：1234.

Tips ： Integers include negative numbers 、 Zero and positive .

Output format ：
Output the Pinyin corresponding to the integer in one line , The Pinyin of each number is separated by a space , There is no final space at the end of the line . Such as
yi er san si.
Examples ：

-600

fu liu ling ling

AC Code ：

#include<iostream>
#include<string>
using namespace std;
#include<stack>
int main()
{
    
    string arr[10]={
    "ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
    int n;
    scanf("%d",&n);
    if(n<0)
    {
    
        printf("%s ","fu");
    }
    stack<int>s;
    int N=abs(n);// Take the absolute value 
    if(N==0)//0 A special sentence is required 
    {
    
        cout<<"ling";
    }
    // Get your numbers , And use the stack to store 
    while(N)
    {
    
        int num=N%10;
        N/=10;
        s.push(num);
    }
    while(!s.empty())
    {
    
        int k=s.top();// Get stack top element 
        // The last one is to judge , There are no spaces 
        if(s.size()==1)
        {
    
            cout<<arr[k];
        }else
        {
    
            cout<<arr[k]<<" ";
        }
        s.pop();// Delete stack top element 
    }
    return 0;
}

1、L1-008 Sum of integral segments fraction 10

Given two integers A and B, Output from A To B All the integers of and the sum of these numbers .

Input format ：
Enter... On one line 2 It's an integer A and B, among −100≤A≤B≤100, Separated by spaces .

Output format ：
First, output sequentially from A To B All integers of , Every time 5 A line of numbers , Each number accounts for 5 Character width , Align right . Finally, in a line, press Sum = X Output the sum of all the numbers X.
Examples ：

-3 8


   -3   -2   -1    0    1
    2    3    4    5    6
    7    8
Sum = 30

AC Code ：

#include<iostream>
int main()
{
    
    int a,b;
    scanf("%d%d",&a,&b);
    int sum=0,cnt=0;
    for(int i=a;i<=b;i++)
    {
    
        printf("%5d",i);// %md  To the right , The character width is m
        cnt++;
        if(cnt%5==0&&b-a!=4)// b-a!=4 Prevent only 5 A situation where a line breaks when a number is added 
        {
    
            printf("\n");
        }
        sum+=i;
    }
    printf("\nSum = %d",sum);// Notice the spaces 
    return 0;
}

1、L1-009 N Sum the numbers fraction 20

The requirement of this question is very simple , Is o N A number and . The trouble is , These numbers are expressed as rational numbers / Given in the form of denominator , The sum you output must also be in the form of a rational number .

Input format ：
The first line of input gives a positive integer N（≤100）. The next line is formatted a1/b1 a2/b2 … give N Rational number . Make sure that all molecules and denominators are in the range of long integers . in addition , The sign of a negative number must be in front of the molecule .

Output format ：
Output the simplest form of the sum of the above numbers —— Write the result as an integer Fraction part , The fraction is written as a molecule / The denominator , The numerator is required to be smaller than the denominator , And they don't have a common factor . If the integral part of the result is 0, Only the fraction part is output .
Examples ：

// Examples 1
5
2/5 4/15 1/30 -2/60 8/3

3 1/3

// Examples 2
2
4/3 2/3

2
// Examples 3
3
1/3 -1/6 1/8

7/24

AC Code ：

#include<iostream>
// Find the greatest common divisor of the numerator and denominator 
int gcd(int a,int b)
{
    
    if(b==0)return a;
    else return gcd(b,a%b);
}
// Use a structure to store scores 
struct Frac
{
    
  long long up,down;// Molecular denominator 
  Frac(){
    }
  Frac(long long u,long long d)
  {
    
      up=u,down=d;
  }
};
// The simplification of fractions 
Frac redu(Frac res)
{
    
// If the denominator is negative , To correct the denominator , Molecules become negative 
    if(res.down<0)
    {
    
        res.up=-res.up;
        res.down=-res.down;
    }
    // The score is 0 Words , Let the molecule be 0, The denominator is 1
    if(res.up==0)
    {
    
        res.down=1;
    }else
    {
    
    // And divide by the greatest common divisor , Simplify fractions 
        int d=gcd(abs(res.up),res.down);
        res.up/=d;
        res.down/=d;
    }
    return res;
}
// Output score 
void showResult(Frac r)
{
    
    r=redu(r);// Simplify first 
    // If it's an integer , Direct output 
    if(r.down==1)printf("%lld\n",r.up);// %lld  Output long integer 
    else if(abs(r.up)>r.down)// If it is a false score 
    {
    
    // Output the integer part first , Then output the rest 
        printf("%lld %lld/%lld\n",r.up/r.down,abs(r.up)%r.down,r.down);
    }else{
    
        printf("%lld/%lld\n",r.up,r.down);
    }
}
// The addition of fractions , Applies to negative fractions 
Frac add(Frac f1,Frac f2)
{
    
    Frac res;
    res.up=f1.up*f2.down+f1.down*f2.up;
    res.down=f1.down*f2.down;
    return redu(res);
}
int main()
{
    
    int N;
    scanf("%d",&N);
    struct Frac arr[N];// Create a structure array to store the scores 
    for(int i=0;i<N;i++)
    {
    
        scanf("%lld/%lld",&arr[i].up,&arr[i].down);
    }
    Frac res(0,1);// Store results 
    for(int i=0;i<N;i++)
    {
    
        res=add(res,arr[i]);// Cumulative sum 
    }
    showResult(res);
    return 0;
}

1、L1-010 Compare the size fraction 10

This question requires any input 3 Integers output from small to large .

Input format :
Enter... On one line 3 It's an integer , Separated by spaces .

Output format :
In one line 3 Integers output from small to large , In the meantime “->” Connected to a .

Examples ：

4 2 8

2->4->8

AC Code ：

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    
    int arr[3];
    for(int i=0;i<3;i++)
    {
    
        scanf("%d",&arr[i]);
    }
    sort(arr,arr+3);// Sort directly into the array 
    printf("%d->%d->%d",arr[0],arr[1],arr[2]);
    return 0;
}

L1-011 A-B fraction 20

This problem requires you to calculate A−B. But the trouble is ,A and B All strings —— That is, from the string A Put the string in B Delete all the included characters , The rest of the characters are strings A−B.

Input format ：
Enter in 2 The string is given one after the other A and B. Neither string is longer than 10
4
, And make sure that every string is made visible ASCII Code and white space characters , End with a line break .

Output format ：
Print out... On one line A−B The result string of .
Examples ：

I love GPLT!  It's a fun game!
aeiou

I lv GPLT!  It's  fn gm!

AC Code ：
edition 1
Mark output method

#include<iostream>
using namespace std;
#include<cstring>
const int n=10010;
int main()
{
    
    char a[n],b[n];
    // Read character array 
    cin.getline(a,sizeof(a));
    cin.getline(b,sizeof(b));
    // Only use strlen(), Will not contain the... At the end of the string '\0'
    for(unsigned i=0;i<strlen(a);i++)
    {
    
        bool flag=true;// Mark whether to output 
        for(unsigned j=0;j<strlen(b);j++)
        {
    
            if(a[i]==b[j])
            {
    
                flag=false;// There is , No output 
                break;
            }
            
        }
        if(flag)
       {
    
          printf("%c",a[i]);
       }
    }
    return 0;
}

edition 2
String built-in function method

#include<iostream>
using namespace std;
#include<string>
int main()
{
    
    string a,b;
    // Read a line of string 
    getline(cin,a);
    getline(cin,b);
    int pos;
    // The length of the string is in length() Method to get 
    for(unsigned i=0;i<b.length();i++)
    {
    
        while((pos=a.find(b[i]))!=-1)
        {
    
            a.erase(pos,1);// Delete pos The character of position 
        }
    }
    cout<<a;
    return 0;
}

（1）erase(pos,n); Remove from pos At the beginning n Characters , such as erase(0,1) Is to delete the first character
（2）erase(position); Delete position One character at (position It's a string Iterator of type )
（3）erase(first,last); Remove from first To last Characters between （first and last They're all iterators ）

L1-012 Calculate the index fraction 5

I really didn't lie to you , This is a simple question —— No more than for any given 10 The positive integer n, You are required to output 2 ^n.
Input format ：
The input gives no more than 10 The positive integer n.

Output format ：
Format in one line 2^n = The result of the calculation is Output 2 ^n Value .
Examples ：

5

2^5 = 32

AC Code ：

#include<iostream>
#include<cmath>
using namespace std;

int main()
{
    
    int n;
    scanf("%d",&n);
    double ans=pow(2.0,n);// Use built-in functions to calculate powers , The return is double type 
    printf("2^%d = %.f",n,ans);
    return 0;
}

L1-013 Calculate factorial sum fraction 10

For a given positive integer N, You need to calculate S=1!+2!+3!+…+N!.

Input format ：
The input gives no more than 10 The positive integer N.

Output format ：
Output in one line S Value .
Examples ：

3

9

AC Code ：

#include<iostream>
// Recursively factoring 
long long fact(int n)
{
    
    if(n==1)return 1;
    else return n*fact(n-1);
}
// Sum of factorials 
long long getSum(int n)
{
    
    long long sum=0;
    while(n)
    {
    
        sum+=fact(n);
        n--;
    }
    return sum;
}
int main()
{
    
    int n;
    scanf("%d",&n);
    printf("%lld",getSum(n));
    return 0;
}

L1-016 Check your ID card fraction 15

A legal ID number is from 17 Location 、 Date number and sequence number plus 1 Bit check code composition . Check code calculation rules are as follows ：

First of all, to the front 17 Bit number weighted sum , The weight is assigned to ：{7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2}; Then the sum of the calculated values is compared to the 11 Take the modulus to get the value Z; Finally, according to the following relationship Z Value and check code M Value ：

Z：0 1 2 3 4 5 6 7 8 9 10
M：1 0 X 9 8 7 6 5 4 3 2

Now give me some ID number. , Please verify the validity of the check code , And output the problem number .

Input format ：
Enter the first line to give a positive integer N（≤100） The number of the ID number entered. . And then N That's ok , Each line gives 1 individual 18 I. D. number .

Output format ：
Output each line in the order of input 1 Question Id number . This is not before the test 17 Is it reasonable , Just before checking 17 Are all digits and last 1 Accurate calculation of bit check code . If all the numbers are OK , The output All passed.
Examples ：

4
320124198808240056
12010X198901011234
110108196711301866
37070419881216001X

12010X198901011234
110108196711301866
37070419881216001X

2
320124198808240056
110108196711301862

All passed

AC Code ：

#include<iostream>
using namespace std;

const int qz[17]={
    7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};// The weight 
const char m[11]={
    '1','0','X','9','8','7','6','5','4','3','2'};// Check code M
// Verify whether the check code is correct 
bool jiaoyan(char id[])
{
    
    int z=0,sum=0;
    for(int i=0;i<17;i++)
    {
    
        sum+=(id[i]-'0')*qz[i];//char First convert to int Calculate again 
    }
    z=sum%11;
    return m[z]==id[17];// Abbreviation 
    /* if(m[z]==id[17]) { return true; }else { return false; } */
}
bool judge(char id[])
{
    
    for(int i=0;i<17;i++)
    {
    
        if(!isdigit(id[i]))// front 17 Bits have a direct that is not a number false
        {
    
            return false;
            break;
        }
    }
    return true;
}
int main()
{
    
    int N;
    scanf("%d",&N);
    char id[20];
    int cnt=0;
    while(N--)
    {
    
        scanf("%s",id);
        bool flag=false;
        if(jiaoyan(id)&&judge(id))
        {
    
            flag=true;
        }
        if(!flag)
        {
    
            cnt++;
            cout<<id<<endl;
        }
    }
    if(cnt==0)//cnt still 0 Explain that there is no mistake 
    {
    
        cout<<"All passed"<<endl;
    }
    return 0;
}

L1-017 How many fraction 15

An integer “ The degree of a second offense ” Defined as contained in the number 2 The ratio of the number of bits to the number of bits . If this number is negative , The degree increases 0.5 times ; If it's still an even number , Is to add 1 times . For example, digital -13142223336 It's a 11 digit , Among them is 3 individual 2, And it's negative , And even , Then the second degree of its offense is calculated as ：3/11×1.5×2×100%, about 81.82%. In this case, you calculate how many dimes a given integer has .

Input format ：
The first line of input gives no more than one 50 An integer N.

Output format ：
Output in one line N The degree of a second offense , Keep two decimal places .
Examples ：

-13142223336

81.82%

AC Code ：

#include<iostream>
#include<cstring>
using namespace std;
double judge(char N[],int len)
{
    
    int isFu= N[0]=='-' ? 1:0;
    char c=N[len-1];
    int oushu= ((c=='0')||(c=='2')||(c=='4')||(c=='6')||(c=='8')) ? 1:0;
    int count=0;
    for(int i=0;i<len;i++)
    {
    
        if(N[i]=='2')count++;
    }
    int cnt= N[0]=='-' ? len-1:len;// If it's negative , The number of digits should be reduced by one 
    double ans=((double)(count)/cnt)*(1+(0.5)*isFu)*(1+1*oushu);
    return ans;
}
int main()
{
    
    char N[55];
    scanf("%s",N);
    int len=strlen(N);
    if(N[0]==0)
    {
    
        printf("0");
    }else
    {
    
        printf("%.2f%%",judge(N,len)*100);// The output percentage sign should be preceded by %
    }
    return 0;
}

L1-018 Big Ben fraction 10

There is a claim on weibo “ Big Ben V” The guy who , Bells are rung every day to urge the farmers to take good care of their health and go to bed early . But since Ben doesn't work and rest regularly on his own , So the ringing of the bell is irregular . Generally, the number of striking bells is determined by the striking time , If I happen to knock on the hour , that “ When ” It's going to be equal to that integral ; If it's past the hour , I'm just going to knock on the next one . in addition , Although one day 24 Hours , The clock struck only the second half of the day 1~12 Next . For example, in 23:00 bell , Namely “ Dang dang dang dang dang dang dang dang dang dang dang dang ”, And by the 23:01 It would be “ Dangdang dangdang dangdang dangdang ”. In the middle of the night 00:00 Until noon 12:00 period （ Endpoint time is included ）, Stupid clocks don't strike .

Now write a program , Ring Big Ben according to the current time .

Input format ：
Enter the first line as follows hh:mm Gives the current time . among hh Is the hour , stay 00 To 23 Between ;mm Is a minute , stay 00 To 59 Between .

Output format ：
Ring Big Ben according to the current time , That is, the corresponding number of output in a line Dang. If it's not the bell-ringing period , The output ：

Only hh:mm.  Too early to Dang.

Examples ：

19:05
DangDangDangDangDangDangDangDang

07:05
Only 07:05.  Too early to Dang.// Two spaces 

AC Code ：

#include<iostream>
using namespace std;

int main()
{
    
    int h,m;
    scanf("%d:%d",&h,&m);
    int flag= m==0 ? 0:1;// Minutes are not 0 Just hit the next point 
    int cnt;// Record the number of strokes 
    if(h<=12)
    {
    
        printf("Only %02d:%02d. Too early to Dang.",h,m);// Notice that there are two spaces 
    }else{
    
        cnt=h-12+flag;// Count times 
        for(int i=1;i<=cnt;i++)
        {
    
            printf("Dang");
        }
    }
    return 0;
}

L1-019 Who first down fraction 15

Boxing is an interesting part of the ancient Chinese wine culture . The method of two people on the wine table is ： Shout out a number from each mouth , Draw a number with your hand . If someone's number is exactly equal to the sum of the two Numbers , Who is lost , The loser gets a glass of wine . Either a win or a loss goes on to the next round , Until the only winner comes along .

So let's give you a 、 B how much for two （ How many cups can you drink at most ） And punching records , Please judge which of the two should go first .

Input format ：
The first line of input gives us a 、 B how much for two （ No more than 100 Non-negative integer ）, Space off . The next line gives a positive integer N（≤100）, And then N That's ok , Each line gives a record of one stroke , The format is ：

 A shout   A stroke   Ethyl shout   B row 

Among them, shouting is the number , A stroke is a number drawn out , No more than 100 The positive integer （ Row with both hands ）.

Output format ：
Output the first person to fall on the first line ：A On behalf of a ,B On behalf of the b . The second line shows how many drinks the person who didn't pour had . They guarantee that one person will fall . Note that the program terminates when someone drops , You don't have to deal with the rest of the data .

sample input ：

1 1
6
8 10 9 12
5 10 5 10
3 8 5 12
12 18 1 13
4 16 12 15
15 1 1 16

A
1

AC Code ：

#include<iostream>
using namespace std;

int main()
{
    
    int A,B;
    scanf("%d%d",&A,&B);
    int N;
    scanf("%d",&N);
    int a,b,c,d;
    int cntA=0,cntB=0;// Number of drinks 
    while(N--)
    {
    
        scanf("%d%d%d%d",&a,&b,&c,&d);
        
        cntA+= a+c==b&&b!=d ? 1:0;// Win without drinking 
        cntB+= a+c==d&&b!=d ? 1:0;
        if(cntA>A)
        {
    
            printf("A\n%d",cntB);
            break;// When one side falls, it stops 
        }else if(cntB>B)
        {
    
            printf("B\n%d",cntA);
            break;
        }
    }
    return 0;
}

L1-021 Say the important thing three times fraction 5

This super simple problem doesn't have any input .

All you have to do is say this very important sentence —— “I’m gonna WIN!”—— Output three times in a row .

Notice that each row is occupied , You cannot have any extra characters except for the carriage return on each line .

I'm gonna WIN!
I'm gonna WIN!
I'm gonna WIN!

for i in range(3):% Generate 0、1、2 Three times 
    print("I'm gonna WIN!")

L1-022 Parity separation fraction 10

Given N A positive integer , Please count the number of odd and even Numbers ？

Input format ：
The first line of input gives us a positive integer N（≤1000）; The first 2 Line is given N Nonnegative integers , Space off .

Output format ：
Output the number of odd Numbers successively in a line 、 Number of even Numbers . In the middle to 1 Space separation .

Examples ：

9
88 74 101 26 15 0 34 22 77

3 6

AC Code ：

import java.util.Scanner;

public class Main
{
    
    public static void main(String args[])
    {
    
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        int cntA=0,cntB=0;
        // Enter one to judge one 
        for(int i=0;i<n;i++)
        {
    
            int num=in.nextInt();
            if(num%2==0)
            {
    
                cntA++;
            }else 
            {
    
                cntB++;
            }
        }
        System.out.printf("%d %d",cntB,cntA);
    }
}

L1-023 Output GPLT fraction 20

A given length does not exceed 10000 Of 、 A string consisting only of English letters . Please reorder the characters , Press GPLTGPLT… Output in this order , And ignore the other characters . Of course , Four types of characters （ Case insensitive ） It doesn't have to be the same number , If a character has been printed out , Then press the remaining characters GPLT Sequential printing , Until all characters are printed .

Input format ：
The input is given in a line with a length not exceeding 10000 Of 、 A non - empty string consisting only of English letters .

Output format ：
Output the sorted string on a line by the title . The question guarantees that the output is not empty .

Examples ：

pcTclnGloRgLrtLhgljkLhGFauPewSKgt

GPLTGPLTGLTGLGLL

AC Code ：

#include<iostream>
using namespace std;
#include<cstring>
int main()
{
    
    char arr[10010];
    cin.getline(arr,sizeof(arr));
    int g=0,p=0,l=0,t=0;
    for(unsigned i=0;i<strlen(arr);i++)
    {
    
        switch((char)arr[i])// The character is automatically converted to int, So turn back 
        {
    
            case 'G':
            case 'g':
                g++;
                break;
            case 'P':
            case 'p':
                p++;
                break;
            case 'L':
            case 'l':
                l++;
                break;
            case 'T':
            case 't':
                t++;
                break;
        }
    }
    while(1)
    {
    
        if(g)
        {
    
            cout<<"G";
            g--;
        }
        if(p)
        {
    
            cout<<"P";
            p--;
        }
        if(l)
        {
    
            cout<<"L";
            l--;
        }
        if(t)
        {
    
            cout<<"T";
            t--;
        }
        if(g==0&&p==0&&l==0&&t==0)break;
    }
    return 0;
}

L1-024 The day after tomorrow fraction 5

If today is Wednesday , The day after tomorrow is Friday ; If today is Saturday , The day after tomorrow is Monday . We use Numbers 1 To 7 Monday to Sunday . Given a day , Please export that day “ The day after tomorrow ” What day is .

Input format ：
The first line of input gives a positive integer D（1 ≤ D ≤ 7）, Represents a day of the week .

Output format ：
Output in one line D What day is the day after tomorrow .

Examples ：

5
7

AC Code ：

#include<iostream>
using namespace std;

int main()
{
    
    int D;
    cin>>D;
    if(D<=5)
    {
    
        cout<<(D+2);
    }else if(D==6)
    {
    
        cout<<1;
    }else if(D==7)
    {
    
        cout<<2;
    }
    return 0;
}

L1-026 I Love GPLT fraction 5

This super simple problem doesn't have any input .

All you have to do is say this very important sentence —— “I Love GPLT”—— Vertical output is OK .

So-called “ Vertical output ”, It's one line per character （ Including Spaces ）, That is, each line can only have 1 Two characters and carriage return .

I
 
L
o
v
e
 
G
P
L
T

AC Code ：

public class Main
{
    
    public static void main(String args[])
    {
    
        String s="I Love GPLT";
        for(int i=0;i<s.length();i++)
        {
    
            System.out.println(s.charAt(i));
        }
    }
}

L1-028 Judgement primes fraction 10

The goal of this question is very simple , Is to determine whether a given positive integer is a prime number .

Input format ：
Enter a positive integer on the first line N（≤ 10）, And then N That's ok , Each line gives a less than 2
31
A positive integer that needs to be judged .

Output format ：
For each positive integer to be judged , If it's a prime , Output in one line Yes, Otherwise output No.

Examples ：

2
11
111

Yes
No

import java.util.Scanner;

public class Main
{
    
    public static void main(String args[])
    {
    
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        for(int i=0;i<n;i++)
        {
    
            int num=in.nextInt();
            if(isPrime(num))
            {
    
                System.out.println("Yes");
            }else 
            {
    
                System.out.println("No");
            }
        }
    }
    public static boolean isPrime(int num)
    {
    
        if(num==1)return false;//1 A special sentence is required 
        int temp=(int)Math.sqrt(num);//sqrt Get is double
        for(int i=2;i<=temp;i++)
        {
    
            if(num%i==0)
            {
    
                return false;
                
            }
        }
        return true;
    }
}

L1-029 Is it too fat fraction 5

It is said that a person's standard weight should be his height （ Company ： centimeter ） subtract 100、 Multiplied by 0.9 The kilogram obtained . It is known that the value of the market weight is twice that of the kilogram . Now give someone height , Please calculate the standard weight ？（ By the way, I'll calculate it for myself ……）

Input format ：
The first line of input gives a positive integer H（100 < H ≤ 300）, For someone's height .

Output format ：
Output the corresponding standard weight in one line , Unit: market Jin , After decimal point 1 position .

Examples ：

169

124.2

#include<iostream>

int main()
{
    
    int n=0;
    scanf("%d",&n);
    double w=2*(n-100)*0.9;
    printf("%.1f",w);
    return 0;
}