正整数A的“DA(为1位整数)部分”定义为由A中所有DA组成的新整数PA。例如:给定A = 3862767,DA = 6,则A的“6部分”PA是66,因为A中有2个6。
现给定A、DA、B、DB,请编写程序计算PA + PB。
输入在一行中依次给出A、DA、B、DB,中间以空格分隔,其中0 < A, B < 1010。
在一行中输出PA + PB的值。
3862767 6 13530293 3
399
A,B,C,D = input().split() E = A.count(B) * B E = eval(E) if E else 0 F = C.count(D) * D F = eval(F) if F else 0 print(E+F)
from collections import Counter #序列计数
def to_num(c,c_num):
num=0
for i in range(c_num):
num=num*10+int(c)
return num
def function():
str1,s1,str2,s2=input().split()
word_counter1=Counter(str1)
c_num1=word_counter1[s1]
word_counter2=Counter(str2)
c_num2=word_counter2[s2]
num1=to_num(s1,c_num1)
num2=to_num(s2,c_num2)
print(num1+num2)
if __name__ == '__main__':
function()
import re import string PA,PB = 0,0 listI = [i for i in input().split()] A = listI[0] DA = listI[1] B = listI[2] DB = listI[3] Da = int(DA) Db = int(DB) pattern = re.compile(DA) numDA = len(pattern.findall(A)) pattern = re.compile(DB) numDB = len(pattern.findall(B)) for i in range(numDA): PA = PA + Da * (10**(i)) for i in range(numDB): PB = PB + Db * (10**(i)) SUM = PA+ PB print(SUM)
a, d_a, b, d_b = input().split() def get_p(digit, d): p = '' for num in digit: if num == d: p += num if p == '': p = 0 else: p = int(p) return p p_a = get_p(a, d_a) p_b = get_p(b, d_b) print(p_a + p_b)
#!/usr/bin/env python #-*- coding:utf8 -*- def findNum(s1, a, s2, b): n1, n2 = 0, 0 for i in s1: if i == a: n1 = n1 *10 + 1 for i in s2: if i == b: n2 = n2 *10 + 1 n1 *= int(a) n2 *= int(b) return n1+n2 if __name__ == '__main__': s1, a, s2, b = raw_input().split() print findNum(s1, a, s2, b)