A reversible prime in any number system is a prime whose “reverse” in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (<10​5​​) and D (1<D≤10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line Yes if N is a reversible prime with radix D, or No if not.

Sample Input:

73 10
23 2
23 10
-2

      
    

Sample Output:

Yes
Yes
No

编译器：Python(python3)

写了三个函数分别是：素数判断，十进制转任意进制的逆序和任意进制转十进制

import math
def IsPrime(num):
    if num == 0 | num == 1:
        return False
    for i in range(2,int(math.sqrt(num))+1):
        if num % i == 0:
            return False
    return True

def reverse_Radix(origin_num,base_num):
    result_num_lsit = []
    if origin_num < base_num:
        return origin_num
    else:
        while True:
            div, mod = divmod(origin_num,base_num)
            # result_num_lsit.insert(0,str(mod))
            result_num_lsit.append(str(mod))
            origin_num = div
            if origin_num < base_num:
                # result_num_lsit.insert(0,str(origin_num))
                result_num_lsit.append(str(origin_num))
                result_num = int(''.join(result_num_lsit))
                return result_num

def ToDecimal(origin_num,base_num):
    result_num = 0
    index = 0
    while True:
        div, mod = divmod(origin_num,10)
        result_num = result_num + mod*base_num**index
        index+=1
        origin_num = div
        if div == 0:
            return result_num

while True:
    Input_Str = input().split(' ')
    if int(Input_Str[0]) < 0:
        break

    origin_num = int(Input_Str[0])
    base_num = int(Input_Str[1])

    reverse_num = reverse_Radix(origin_num,base_num)

    result_num = ToDecimal(reverse_num,base_num)

    if IsPrime(origin_num) & IsPrime(result_num):
        print('Yes')
    else:
        print('No')

Given three integers A, B and C in [−2​63​​,2​63​​], you are supposed to tell whether A+B>C.

Input Specification:

The first line of the input gives the positive number of test cases, T (≤10). Then T test cases follow, each consists of a single line containing three integers A, B and C, separated by single spaces.

Output Specification:

For each test case, output in one line Case #X: true if A+B>C, or Case #X: false otherwise, where X is the case number (starting from 1).

Sample Input:

3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0

      
    

Sample Output:

Case #1: false
Case #2: true
Case #3: false

编译器：Python(python3)

判断溢出

N = int(input())

index = 1
while(index<=N):
    flag = 0
    a, b, c = map(int,input().split(' '))
    if a>0 & b>0 & a+b<0:
        flag = 1
    elif a<0 & b<0 & a+b>=0:
        flag = 0
    elif a+b>c:
        flag = 1
    print('Case #{}:'.format(index),"true" if flag else "false")
    index +=1

Excel can sort records according to any column. Now you are supposed to imitate this function.

Input Specification:

Each input file contains one test case. For each case, the first line contains two integers N (≤10​5​​) and C, where N is the number of records and C is the column that you are supposed to sort the records with. Then N lines follow, each contains a record of a student. A student’s record consists of his or her distinct ID (a 6-digit number), name (a string with no more than 8 characters without space), and grade (an integer between 0 and 100, inclusive).

Output Specification:

For each test case, output the sorting result in N lines. That is, if C = 1 then the records must be sorted in increasing order according to ID’s; if C = 2 then the records must be sorted in non-decreasing order according to names; and if C = 3 then the records must be sorted in non-decreasing order according to grades. If there are several students who have the same name or grade, they must be sorted according to their ID’s in increasing order.

Sample Input 1:

3 1
000007 James 85
000010 Amy 90
000001 Zoe 60

      
    

Sample Output 1:

000001 Zoe 60
000007 James 85
000010 Amy 90

      
    

Sample Input 2:

4 2
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 98

      
    

Sample Output 2:

000010 Amy 90
000002 James 98
000007 James 85
000001 Zoe 60

      
    

Sample Input 3:

4 3
000007 James 85
000010 Amy 90
000001 Zoe 60
000002 James 90

      
    

Sample Output 3:

000001 Zoe 60
000007 James 85
000002 James 90
000010 Amy 90

编译器：Python(python3)

这题用list的sort方法，其中可以按key关键字排序，但是python的缺点就是最后一个测试点超时，但是不想用C++再写了。

N, ref = map(int, input().split())

records = []

for i in range(N):
    s_id, name, score = input().split()
    records.append([s_id, name, score])

records.sort(key=lambda x: (x[ref - 1],x[0]))
for i in records:
    print(" ".join(i))

Given any string of N (≥5) characters, you are asked to form the characters into the shape of U. For example, helloworld can be printed as:

h  d
e  l
l  r
lowo

      
    

That is, the characters must be printed in the original order, starting top-down from the left vertical line with n​1​​ characters, then left to right along the bottom line with n​2​​ characters, and finally bottom-up along the vertical line with n​3​​ characters. And more, we would like U to be as squared as possible — that is, it must be satisfied that n​1​​=n​3​​=max { k | k≤n​2​​ for all 3≤n​2​​≤N } with n​1​​+n​2​​+n​3​​−2=N.

Input Specification:

Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

Output Specification:

For each test case, print the input string in the shape of U as specified in the description.

Sample Input:

helloworld!

      
    

Sample Output:

h   !
e   d
l   l
lowor

编译器：Python(python3)

string = input()
N = len(string)
height = (N+2)//3
width = N-2*height-2
for i in range(height - 1):
    print(string[i] + ' '*(width+2) + string[-i-1])
print(string[height-1:N-height+1])

This time, you are supposed to find A×B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:

K N​1​​ a​N​1​​​​ N​2​​ a​N​2​​​​ … N​K​​ a​N​K​​​​

where K is the number of nonzero terms in the polynomial, N​i​​ and a​N​i​​​​ (i=1,2,⋯,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10, 0≤N​K​​<⋯<N​2​​<N​1​​≤1000.

Output Specification:

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

Sample Input:

2 1 2.4 0 3.2
2 2 1.5 1 0.5

      
    

Sample Output:

3 3 3.6 2 6.0 1 1.6

编译器：Python(python3)

p1 = list(input().split())
p2 = list(input().split())

d1 = {}
d2 = {}
d3 = {}


list1 = []


for i in range(1,len(p1),2):
    if float(p1[i+1]) !=0:
        d1[int(p1[i])] = float(p1[i+1])
for i in range(1,len(p2),2):
    if float(p2[i+1]) !=0:
        d2[int(p2[i])] = float(p2[i+1])
        
for k1 in d1:
    for k2 in d2:
        temp = d3.get(k1+k2,0) + d1.get(k1)*d2.get(k2)
        if temp !=0:
            d3[k1+k2] = round(temp,1)
        else:
            del d3[k1+k2]
list1 = [str(len(d3))]            
if len(d3) != 0:            
    for i in sorted(d3.keys(),reverse = True):
        list1.append(str(i))
        list1.append(str(d3[i]))
else:
    list1.append(str(0))
    list1.append(str(0))

print(' '.join(list1))