2014
02-14

# ISBN

Farmer John’s cows enjoy reading books, and FJ has discovered that his cows produce more milk when they read books of a somewhat intellectual nature. He decides to update the barn library to replace all of the cheap romance novels with textbooks on algorithms and mathematics. Unfortunately, a shipment of these new books has fallen in the mud and their ISBN numbers are now hard to read.

An ISBN (International Standard Book Number) is a ten digit code that uniquely identifies a book. The first nine digits represent the book and the last digit is used to make sure the ISBN is correct. To verify that an ISBN number is correct, you calculate a sum that is 10 times the first digit plus 9 times the second digit plus 8 times the third digit … all the way until you add 1 times the last digit. If the final number leaves no remainder when divided by 11, the code is a valid ISBN.

For example 0201103311 is a valid ISBN, since
10*0 + 9*2 + 8*0 + 7*1 + 6*1 + 5*0 + 4*3 + 3*3 + 2*1 + 1*1 = 55.

Each of the first nine digits can take a value between 0 and 9. Sometimes it is necessary to make the last digit equal to ten; this is done by writing the last digit as X. For example, 156881111X is a valid ISBN number.

Your task is to fill in the missing digit from a given ISBN number where the missing digit is represented as ‘?’.

* Line 1: A single line with a ten digit ISBN number that contains ‘?’ in a single position

* Line 1: A single line with a ten digit ISBN number that contains ‘?’ in a single position

15688?111X

1

# include <stdio.h>

int main ()
{
char str[20] ;
int sum, i, pos ;
while (gets(str))
{
sum = 0 ;
for (i = 0 ; str[i] ; i++)
{
if (str[i] == '?') pos = i ;
else if (str[i] == 'X') sum += 10 ;
else sum += (10-i) * (str[i]-'0') ;
}
if (pos == 9)
{
if ((sum+10)%11==0) puts ("X") ;
else{
for (i = 0 ; i <= 9 ; i++){
if ((sum + i)%11==0){ printf ("%d\n", i) ;
break ;}
}
}
}
else
{
for (i =0 ; i <= 9 ; i++)
{
if ((sum + (10-pos)*i)%11 == 0)
{
printf("%d\n", i) ;
break ;
}
}
if (i >=10) puts ("-1") ;
}
}
return 0 ;
}

1. #!/usr/bin/env python
def cou(n):
arr =
i = 1
while(i<n):
arr.append(arr[i-1]+selfcount(i))
i+=1
return arr[n-1]

def selfcount(n):
count = 0
while(n):
if n%10 == 1:
count += 1
n /= 10
return count

2. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
O(n) = O(n-1) + O(n-2) + ….
O(n-1) = O(n-2) + O(n-3)+ …
O(n) – O(n-1) = O(n-1)
O(n) = 2O(n-1)