Problem:


Score : 100 points

Problem Statement

Compute A \times B.

Constraints

1 \leq A \leq 100

1 \leq B \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the value A \times B as an integer.

Sample Input 1

2 5

Sample Output 1

10

We have 2 \times 5 = 10.

Sample Input 2

100 100

Sample Output 2

10000



Sample input:
2 5


Sample output:
10

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-
import sys

import fractions
import copy
import bisect
import math
import numpy as np
import itertools
from itertools import combinations_with_replacement
#import math#数学的計算はこれでいける。普通に0.5乗しても計算可能
#w=input()
from operator import itemgetter
from sys import stdin
#input = sys.stdin.readline#こっちの方が入力が早いが使える時に使っていこう
from operator import mul
from functools import reduce
from collections import Counter
#from collections import deque

#input = stdin.readline


j=0
k=0
n=3
r=1
a=[0]
#n=int(input())
#r=int(input())
#print(M)
#A=int(input())
#B=int(input())
#print(N)
"1行1つの整数を入力を取得し、整数と取得する"


#number_list=list(map(int, input().split(" ")))#数字の時
#print(number_list)
"12 21 332 とか入力する時に使う"
"1行に複数の整数の入力を取得し、整数として扱う"

'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
#メモ
for i in number_list:#こっちの方がrage使うより早いらしい
    print(number_list[i-1])#

'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''

'''
x=[]
y=[]
for i in range(N):
    x1, y1=[int(i) for i in input().split()]
    x.append(x1)
    y.append(y1)
print(x)
print(y)
"複数行に2数値を入力する形式 x座標とy座標を入力するイメージ"
'''

'''
mixlist=[]
for i in range(N):
    a,b=input().split()
    mixlist.append((int(a),b))
print(mixlist)
"複数行にintとstrを複合して入力するやつ，今回はリスト一つで処理している"
'''

'''
#array=[input().split()for i in range(N)]
#print(type(array[0][0]))
#print(array)
"正方行列にstr型の値を入力"
'''

#brray=[list(map(int, input().split(" ")))for i in range(N)]
#print(type(brray[0][0]))
#print(brray)
'''
入力
1 2
4 5
7 8
出力結果
[[1, 2], [4, 5], [7, 8]]
'''
"列数に関して自由度の高いint型の値を入力するタイプの行列"
#以下に別解を記載
#N, M = [int(i) for i in input().split()]
'''
table = [[int(i) for i in input().split()] for m in range(m)]

print(type(N))
print(N)
print(type(M))
print(M)

print(type(table))
print(table)
'''




#s=input()
#a=[int(i) for i in s]
#print(a[0])
#print([a])
#単数値．桁ごとに分割したい．入力と出力は以下の通り
#イメージとして1桁ごとにリストに値を入れているかんじ
#intを取ると文字列分解に使える
'''
入力
1234
出力
1
[[1, 2, 3, 4]]
'''




'''
word_list= input().split(" ")
print(word_list[0])
"連続文字列の入力"
"qw er ty とかの入力に使う"
"入力すると空白で区切ったところでlistの番号が与えられる"
'''



'''
A, B, C=stdin.readline().rstrip().split()#str style 何個でもいけることが多い
print(A)
"リストではなく独立したstr型を入れるなら以下のやり方でOK"
'''


#a= stdin.readline().rstrip()
#print(a.upper())
"aという変数に入っているものを大文字にして出力"


#a,b=map(int, input().split())  #int style 複数数値入力　「A B」みたいなスペース空いた入力のとき

#なんかうまく入力されるけど
#a=[[int(i) for i in 1.strip()]for 1 in sys.stdin]
#a = [[int(c) for c in l.strip()] for l in sys.stdin]]
#print(a)
#複数行の数値を入力して正方行列を作成


##############################################################################################
##############################################################################################
#under this line explains example calculation



'''
コンビネーションの組み合わせの中身を出力する形式
for i in itertools.combinations(brray, 2)
combinationsをpermutationsにすれば順列になる
今回なら(abc133B)
入力
1 2
5 5
-2 8
出力
[[1, 2], [5, 5], [-2, 8]]
もちろん一次元リストでも使えるし
何よりiもリストのように使えるので便利
'''

#nCr combination
'''
def cmb(n,r):
    #When n < r , this function isn't valid
    r= min(n-r,r)
    #print(n,r)
    if r == 0: return 1
    over = reduce(mul, range(n, n-r, -1))
    #flochart mul(n,n-1)=x
    #next     mul(x,n-2)........(n-r+1,n-r)
    #mul read a,b and returns a*b

    under = reduce(mul, range(1, r+1))
    #print(over, under)
    #reduce is applied mul(1,2)=2
    #next   mul(2,3)=6
    #next   mul(6,4)=4.........last(r!,r+1)=r+1!

    return over // under
    #// is integer divide
    #calc example 5C2
    #over=5*4*3
    #under=3*2*1
a = cmb(n, r)
#print(a)
'''

'''
import itertools
from itertools import combinations_with_replacement

combinationについて
以下の違いを意識しよう
combinations() p, r 長さrのタプル列、ソートされた順で重複なし

combinations_with_replacement() p, r 長さrのタプル列、ソートされた順で重複あり


使用例 　　　　　　　　　　　                  出力　　
combinations('ABCD', 2)　　　　　　　　　　　　AB AC AD BC BD CD
combinations_with_replacement('ABCD', 2)　　 AA AB AC AD BB BC BD CC CD DD
'''



'''
#集計
#example
#a=[2,2,2,3,4,3,1,2,1,3,1,2,1,2,2,1,2,1]
#a=Counter(a)
for i in a.most_common(n):print(i)
#most_common()メソッドは、出現回数が多い要素順にCounterオブジェクトを並び替えます。
#引数にint型の数字nを設定した場合は、出現回数が高い上位ｎ個の要素を返します。
#何も設定しなければ、コンテナ型にあるすべての要素を出現回数の順番に並び替えたタプル型オブジェクトを返します。
#out put
#(2, 8)
 #(1, 6)
#(3, 3)
#(4, 1)
'''


#二部探索(binary search)
#A = [1, 2, 3, 3, 3, 4, 4, 6, 6, 6, 6]
#print(A)
#index = bisect.bisect_left(A, 5) # 7 最も左(前)の挿入箇所が返ってきている
#A.insert(index, 5)
#print(index)
#print(A)
'''
bisect.bisect_left(a, x, lo=0, h=len(a))
引数
a: ソート済みリスト
x: 挿入したい値
lo: 探索範囲の下限
hi: 探索範囲の上限
(lo, hiはスライスと同様の指定方法)
bisect_leftはソートされたリストaに対して順序を保ったままxを挿入できる箇所を探索します。leftが示す通り、aにすでにxが存在している場合は、挿入箇所は既存のxよりも左側になります。また、lo, hiを指定することで探索範囲を絞り込むことも可能です。デフォルトはaの全体が探索対象です。
'''

'''
素数の判定
'''
def is_prime(n):
    if n == 1: return False

    for k in range(2, int(np.sqrt(n)) + 1):
        #sqrt(n)+1以上は考えて約数はないので却下
        if n % k == 0:
            return False
            #割り切れたらFalse
    return True


'''
npのmaxとmaximumの違い
xs = np.array([1, -2, 3])
np.max(xs, 0)
この出力は3となります．[1, -2, 3]と0の4つの数字のうち，最も大きい値を出力します．

一方で，[max(1, 0), max(-2, 0), max(3, 0)]を出力したい時があります．
その時は，numpyのmaximum関数を用います．

xs = np.array([1, -2, 3])
np.maximum(xs, 0)  # [1, 0, 3]

'''

def factorial(x, mod=10**9+7):
    # x!
    # 階乗
    # ex) factorial(5) = 120
    tmp = 1
    for i in range(1, x+1):
        tmp = (tmp * i) % mod
    return tmp
    '''
    Basically, when you do a**d % n, you actually have to calculate a**d, which could be quite large.
    But there are ways of computing a**d % n without having to compute a**d itself, and that is what pow does.
    '''
"#a,bの最大公約数"
def gcd(a, b):
    while b:#0以外なら進んでいくが、0になった時にとまる余りが0より、最大公約数と言える
        #print(a,b)#2,3
        #やり方てきにユークリッドの互除法か
        a, b = b, a % b#aには2 bには2%3が入る
    return a

"#a,bの最小公倍数"
def lcm(a, b):
    return a * b // gcd (a, b)

"#nの約数を全て求める"
def divisor(n):
    i = 1
    table = []
    while i * i <= n:
        if n%i == 0:
            table.append(i)
            table.append(n//i)
        i += 1
    table = list(set(table))
    return table


########################################################################
########################################################################
#b2=a[:]  #1次元のときはコピーはこれで良い
#print(b2)

#a= [[0]*3 for i in range(5)] #2次元配列はこう準備、[[0]*3]*5だとだめ
#b3=copy.deepcopy(a)  #2次元配列はこうコピーする
#print(b3)

'''
ascii_letters = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
ascii_lowercase = 'abcdefghijklmnopqrstuvwxyz'
ascii_uppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
digits = '0123456789'
hexdigits = '0123456789abcdefABCDEF'
letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv...\xaf\xb0...
lowercase = 'abcdefghijklmnopqrstuvwxyz'
octdigits = '01234567'
printable = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTU...
punctuation = '!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'
uppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
whitespace = '\t\n\x0b\x0c\r '
'''
'''
引数endに任意の文字列を指定すると、その文字列が改行の代わりに最後に挿入される。例えば連続するprint()の出力を改行なしで単純に連結したい場合は空文字列''を指定すればよい。

print('abc', end='---')
print('xyz')
# abc---xyz

print('abc', end='')
print('xyz')
# abcxyz
source: print_basic.py
なお、デフォルトで改行されるのは引数endのデフォルト値が改行文字'\n'に設定されているから。
'''

def main():
    n=3
    steps=0
    state=0
    w=1
    j=0
    k=0
    #lenを変数に置くと事故るから注意
    #n=int(input())
    #a,b=input().split(" ")#a,bがともに文字列のとき
    a,b=map(int, input().split(" "))
    print(a*b)
    #mapでとってもstrで変換すればappend可能
    #s=input()
    #a=[i for i in s]#int(i)にすれば数字も行ける
    #nl=list(map(int, input().split(" ")))#数字の時
    #print(a)
    #brray=[list(map(int, input().split(" ")))for i in range(n)]
    #print(brray)


    #for i in range(1,n+1,2):#これで1個飛ばしループになる
    # pass
     #print(i)

    l = ['aaa', 'bbb', 'ccc']
    x=l.index('aaa')#ある要素のインデックス番号を取得
    #s = ''.join(l)
    #s = " ".join(map(str,a))#スペースを空ける時はこう
    #print(s)
    # aaabbbccc

    #if state in known_states and not flag: # state => 2 steps => 5
        #こうやってある要素がリストの中にあるか探すのか
    #print(f'steps: {steps}, state: {state}')
    #こう書いてあるとデバックしやすい

    #dp= [[0]*3 for i in range(5)]#列　行
    #dp用の0の入ったやつ
    #dp= [[0]*(w+1) for i in range(n+1)]#0からwまでのw+1回計算するから
    #print(dp)＃初期条件が入る分計算回数+1列分必要(この場合は判断すべきものの数)
    DP = np.zeros(w+1, dtype=int)#これでも一次元リストが作れる
    exdp=np.zeros((3,4)) # ３×４の２次元配列を生成。2次元ならこう
    #dtypeは指定しないとfloatになる
    #for i in range(n):#ちょっとした入力に便利
    #    a, b = map(int, input().split())
        #dp[i][0] += [a]
        #これだとintとlistをつなぐことになって不適
    #    dp[i] += [a]
    #    dp[i] += [b]
        #これはうまくいく
        #やり方はいろいろあるということ
    #print(dp)


    #brray=[list(map(int, input().split(" ")))for i in range(N)
    #print(brray)



    '''
    入力
    1234
    出力
    [1, 2, 3, 4]
    '''
    pin_l=["x" for i in range(10)]#内包表記に慣れろ
    #print(pin_l)

    ls = ["a", "b", "c", "d", "e"]


    #print(ls[2:5])
    #スライスでの取得

    #print(ls[:-3])
    #一番左端から右から3番目より左まで取得

    #print(ls[:4:2])
    #スライスで1個飛ばしで取得
    #ないときは左端スタート
    #始点のインデックス番号 : 終点のインデックス番号 : スキップする数+1
    #print(ls[::2])
    '''
    lsというリストの場合に、1つ飛びの値を取得したい場合には
    ls[::2]
    のようにします。こうすると、
    ["a", "c", "d"]と出力される
    '''

if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
i = list(map(int, input().split()))

print(i[0]*i[1])

Submission status: Accepted

Solution code:
#!/usr/bin/env python3

# import sys
# import math
# sys.setrecursionlimit(10**8)
# INF = float('inf')
# PI = math.pi()
# from functools import lru_cache
# @lru_cache(maxsize=None)

# n = int(input())
a, b = map(int, input().split())
# c = list(map(int, input().split()))
# s = [list(map(int,list(input()))) for i in range(h)]

print(a*b)

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-

###
# main
if(__name__ == '__main__'):
    # input
    a,b = map(int, input().split(" "))

    ans = a * b

    print(ans)

# else:
    # do nothing

Submission status: Accepted

Solution code:
A, B = input().split(' ')

A = int(A)
B = int(B)

print(A * B)

Submission status: Accepted

Solution code:
a,b=[int(i) for i in input().split()]
print(a*b)

Submission status: Accepted

Solution code:
a, b = list(map(int,input().split()))
print(a*b)

Submission status: Accepted

Solution code:
import sys
max_int = 1000000001  # 10^9+1
min_int = -max_int

a, b = map(int, sys.stdin.readline().split())

print(a*b)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
import math
import sys

input = sys.stdin.readline
#n = int(input())
n,k = map(int,input().split())
#A = list(map(int,input().split()))
#s=input()
print(n*k)

Submission status: Accepted

Solution code:
## necessary imports
# import sys
# input = sys.stdin.readline
# from math import ceil, floor;

# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp

## gcd function
def gcd(a,b):
    if a == 0:
        return b
    return gcd(b%a, a)

## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k 
    res = 1
    for i in range(k): 
        res = res * (n - i) 
        res = res / (i + 1) 
    return int(res) 

## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
    hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        if a[mid] < x:
            lo = mid+1
        else:
            hi = mid
    return lo

## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes

## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 

## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b

# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];

# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;


# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x

## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime

#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array

################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve() 
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)

## taking integer array input
def int_array():
    return list(map(int, input().strip().split()))
## taking string array input
def str_array():
    return input().strip().split();

#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');

################### ---------------- TEMPLATE ENDS HERE ---------------- ###################

a,b = int_array();
print(a*b);

Submission status: Accepted

Solution code:
import sys, random, math
from bisect import bisect_left, bisect_right
input = sys.stdin.readline

def main():
    inf = 10 ** 20

#    t = int(input())
#    t, a, b = map(int, input().split())
    t = 1
    
    for _ in range(1, t+1):
    #    print("Case #{}: ".format(_), end = '')
        
        a, b = map(int, input().split())
        print(a*b)
        
        
main()

Submission status: Accepted

Solution code:
Line = [int(s) for s in input().split()]

print(Line[0]*Line[1])

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
import sys
import numpy as np


def solve(A: int, B: int):
    print(A*B)
    return


# Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    A = int(next(tokens))  # type: int
    B = int(next(tokens))  # type: int
    solve(A, B)

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
a,b = map(int, input().split())
print(str(a*b))

Submission status: Accepted

Solution code:
A, B = (int(x) for x in input().split())

print(A*B)

Submission status: Accepted

Solution code:
x1,x2 = [int(s) for s in input().split(" ") ]

print (x1*x2)

Submission status: Accepted

Solution code:
def solve(a, b):
    return a * b


def main():
    a, b = map(int, input().split())
    res = solve(a, b)
    print(res)


def test():
    assert solve(2, 5) == 10
    assert solve(100, 100) == 10000


if __name__ == "__main__":
    test()
    main()

Submission status: Accepted

Solution code:
import sys
import copy
from collections import deque
stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  # ignore trailing spaces

a, b = na()
print(a*b)

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-

a, b = (int(x) for x in input().split())

print(a * b)

Submission status: Accepted

Solution code:
#a
a,b = list(map(int,input().split()))
print(a*b)

Submission status: Accepted

Solution code:
def solve():
    a, b = map(int, input().split())
    
    

    return a*b
    

print(solve())

Submission status: Accepted

Solution code:
a ,b = map(int, input().strip().split())
print(a*b)

Submission status: Accepted

Solution code:
ii = lambda : int(input())
mi = lambda : map(int,input().split())
li = lambda : list(map(int,input().split()))

a,b = mi()
print(a*b)

Submission status: Accepted

Solution code:
a,b = map(int,raw_input().split(' '))
print a*b

Submission status: Accepted

Solution code:
A,B=map(int, input().split())
print(A*B)

Submission status: Accepted

Solution code:
a, b = map(int, input().split(" "))
print( a * b)

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline


def main():
    A, B = map(int, input().split())
    print(A*B)


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
A, B= list(map(int, input().split()))
print(A*B)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
a, b = map(int, input().split())
print(a*b)

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-
"""
Created on Sun May 31 20:59:05 2020

@author: NEC-PCuser
"""

A, B = map(int, input().split())

print(A * B)

Submission status: Accepted

Solution code:
# Contest No.: ABC169
# Problem No.: A
# Solver:      JEMINI
# Date:        20200531

import sys
import heapq

def main():
    a, b = map(int, sys.stdin.readline().split())
    print(a * b)
    return

if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
def solve(a,b):
    return a*b

if __name__ == '__main__':
    a,b = map(int,input().split())
    print(solve(a,b))

Submission status: Accepted

Solution code:
from math import factorial as fact
import math
import sys
import itertools
import numpy as np
from collections import Counter
import datetime
from collections import deque


#入力:N(int:整数)
def input1():
	return int(input())

#入力:N,M(int:整数)
def input2():
	return map(int,input().split())

#入力:[n1,n2,...nk](int:整数配列)
def input_array():
	return list(map(int,input().split()))

def keta(N):
	return len(str(N))


def input_daytime(input_time):
	time,distance=input_time.split()
	t=time.split(":")
	se=str(t[-1]).split(".")
	dt1 = datetime.timedelta(hours=int(t[0]),minutes=int(t[1]),seconds=int(se[0]),milliseconds=int(se[1]))
	return dt1

def combinations(n,r):
	return list(itertools.combinations(range(n),r))



def all_sets(num_list):
	subsets=[]
	for i in range(2,len(num_list) + 1):
		for c in combinations(num_list, i):
			subsets.append(c) 
	return subsets

#公約数の列挙
def ch(x1):
	cf=[]
	for i in range(2,math.sqrt(x1)+1):
		if x1 % i==0:
			cf.append(i)
	return cf

def CountOneRoots(x,y):
	#x:縦，y:横
	return fact(x+y)/fact(x)/fact(y)

# 素因数分解
def prime_factorize(n):
    a = []
    while n % 2 == 0:
        a.append(2)
        n //= 2
    f = 3
    while f * f <= n:
        if n % f == 0:
            a.append(f)
            n //= f
        else:
            f += 2
    if n != 1:
        a.append(n)
    return a
	
def count_str(m):
	tmp={}
	for i in range(10):
		if i in tmp:
			tmp[i]=1
		else:
			tmp[i]+=1
	return tmp

def is_prime(n):
	if n==1: return True

	for i in range(2,int(math.sqrt(n))+1):
		if n%i==0:
			return False
	return True

a,b=input2()
print(a*b)

Submission status: Accepted

Solution code:
m, n = map(int, input().split())
print(m*n)

Submission status: Accepted

Solution code:
if __name__ == "__main__":

     A, B = map(int,input().split()) 
     print(A*B)

Submission status: Accepted

Solution code:
#!/usr/bin/env python
# -*- coding: utf-8 -*-

import sys
sys.setrecursionlimit(10**7)
from pprint import pprint as pp
from pprint import pformat as pf
# @pysnooper.snoop()
#import pysnooper # debug

import math
#from sortedcontainers import SortedList, SortedDict, SortedSet # no in atcoder
import bisect


if __name__ == '__main__':
    a, b = list(map(int, input().split()))
    print(a * b)

    #print('\33[32m' + 'end' + '\033[0m') # debug

Submission status: Accepted

Solution code:
from sys import stdin


def main():
    input = lambda: stdin.readline()[:-1]
    A, B = map(int, input().split())
    print(A * B)


main()

Submission status: Accepted

Solution code:
import itertools
import collections
from collections import Counter
def li():
    return list(map(int,input().split()))
#n=int(input())
p=li()
print(p[0]*p[1])
#for i in range(n):

Submission status: Accepted

Solution code:
A,B = input().split(" ")

print(int(A)*int(B))

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline
def ii(): return int(input())
def mi(): return map(int, input().rstrip().split())
def lmi(): return list(map(int, input().rstrip().split()))
def li(): return list(input().rstrip())
def debug(*args, sep=" ", end="\n"): print("debug:", *args, file=sys.stderr, sep=sep, end=end) if not __debug__ else None
def exit(*arg): print(*arg); sys.exit()
# template

def main():
    a, b = mi()
    print(a * b)
    return

if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
A, B = [int(v) for v in input().strip().split(" ")]

print(A*B)

Submission status: Accepted

Solution code:
def main():
    A,B = map(int,input().split())
    C = A*B
    print(C)
    
if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import sys
sys.setrecursionlimit(10000000)
input=sys.stdin.readline
mod = 10 ** 9 + 7

n,k = map(int,input().split())
print(n*k)

Submission status: Accepted

Solution code:
def main():
    A,B=map(int, input().split())
    print(A*B)
    return

if __name__=='__main__':
    main()

Submission status: Accepted

Solution code:
from sys import stdin, stdout
n, m = map(int, stdin.readline().strip().split())
print(n*m)

Submission status: Accepted

Solution code:
def solve():
    A, B = map(int,input().split())
    print(A*B)

if __name__ == '__main__':
    solve()

Submission status: Accepted

Solution code:
# coding: utf-8
a,b = map(int, input().split())
print(a*b)

Submission status: Accepted

Solution code:
def main():
    a, b = map(int,input().split())
    r = a * b
    print(int(r))

if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
A,B = (int(x) for x in input().split())
print(int(A*B))

Submission status: Accepted

Solution code:
# 2020/05/31
# AtCoder Beginner Contest 169 - A

# Input
a, b = map(int,input().split())

# Calc
ans = a * b

# Output
print(ans)

Submission status: Accepted

Solution code:
import os
import heapq
import sys
import math
import operator
from collections import defaultdict
from io import BytesIO, IOBase
# def gcd(a,b):
#     if b==0:

#         return a
#     else:
#         return gcd(b,a%b)
def main():
    # mod=10**9+7
    #for _ in range(int(input())):
    a,b=map(int,input().split())
    print(a*b)        







BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
a , b = map(int , input().split())

print(a*b)

Submission status: Accepted

Solution code:
A, B = map(int, input().split())

ans = A * B

print(str(ans))

Submission status: Accepted

Solution code:
import sys

readline = sys.stdin.readline
readall = sys.stdin.read
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prn = lambda x: print(*x, sep='\n')

a, b = nm()
print(a*b)

Submission status: Accepted

Solution code:
a, b = (int(i) for i in input().split())
print(a*b)

Submission status: Accepted

Solution code:
A, B = list(map(int, input().split()))


print(A * B)

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
a, b = map(int, input().split())
print(a * b)

Submission status: Accepted

Solution code:
def main():
  A, B = map(int, input().split())
  print(A*B)
if __name__ == "__main__":
  main()

Submission status: Accepted

Solution code:

#!/usr/bin/python
# -*- coding: utf-8 -*-

A,B=map(int,input().split())

print(A*B)

Submission status: Accepted

Solution code:

def parse():
  return map(int, input().split(" "))

def main():
  A, B = parse()
  print(A * B)

if __name__ == "__main__":
  main()

Submission status: Accepted

Solution code:
from sys import stdin, stdout
import heapq
import cProfile
from collections import Counter, defaultdict, deque
from functools import reduce
import math


def get_int():
    return int(stdin.readline().strip())


def get_tuple():
    return map(int, stdin.readline().split())


def get_list():
    return list(map(int, stdin.readline().split()))



def solve():
    n,m = get_tuple()
    print(n*m)



def main():
    solve()


TestCases = False

if TestCases:
    for i in range(get_int()):
        main()
else:
    main()

Submission status: Accepted

Solution code:
#from collections import deque,defaultdict
from sys import stdin
input = stdin.readline
printn = lambda x: print(x,end='')
inn = lambda : int(input())
inl   = lambda: list(map(int, input().split()))
inm   = lambda:      map(int, input().split())
ins = lambda : input().strip()
DBG = True # and False
BIG = 10**18
R = 10**9 + 7

def ddprint(x):
  if DBG:
    print(x)

a,b = inm()
print(a*b)

Submission status: Accepted

Solution code:
x, y= map(int, input().split())
print(x*y)

Submission status: Accepted

Solution code:
a,b = list(map(int, input().split(' ')))
print(a*b)

Submission status: Accepted

Solution code:
a, b =map(int, input().split())
print(a * b)

Submission status: Accepted

Solution code:
def main():
    a, b = map(int, input().split())
    ans = a * b
    print(ans)


if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
 
def main():
    a, b = map(int, readline().split())
    print(a*b)
if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
import sys

A,B=list(map(int,input().split()))

print(A*B)

Submission status: Accepted

Solution code:
import sys
# sys.setrecursionlimit(pow(10, 6))
# sys.stdin = open("input.txt", "r")
# sys.stdout = open("output.txt", "w")
mod = pow(10, 9) + 7
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def l(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]


a, b = sp()
out(a * b)

Submission status: Accepted

Solution code:
def readInt():
    return list(map(int, input().split()))
    
a,b = readInt()
print(a*b)

Submission status: Accepted

Solution code:
A, B = map(int,input().split())

ans = A * B
print(ans)

Submission status: Accepted

Solution code:
import sys
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))  #空白あり
def LI2(): return list(map(int,sys.stdin.readline().rstrip()))  #空白なし
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())  #空白あり
def LS2(): return list(sys.stdin.readline().rstrip())  #空白なし

A,B = map(int,S().split())

print(A*B)

Submission status: Accepted

Solution code:

n , m = map(int , input().split())
print(n*m)