BestCoder网址：http://bestcoder.hdu.edu.cn/contests/contest_show.php?cid=530
A：http://acm.hdu.edu.cn/showproblem.php?pid=4956
B：http://acm.hdu.edu.cn/showproblem.php?pid=4957
C：http://acm.hdu.edu.cn/showproblem.php?pid=4958
D：http://acm.hdu.edu.cn/showproblem.php?pid=4959
A - Poor Hanamichi
Description
Hanamichi is taking part in a programming contest, and he is assigned to solve a special problem as follow: Given a range [l, r] (including l and r), find out how many numbers in this range have the property: the sum of its odd digits is smaller than the sum of its even digits and the difference is 3.
A integer X can be represented in decimal as:
$X = A_n \times 10^n + A_{n-1} \times 10^{n-1} + \ldots + A_2 \times 10^2 + A_1 \times 10^1 + A_0$
The odd dights are $A_1, A_3, A_5 \ldots$ and $A_0, A_2, A_4 \ldots$ are even digits.
Hanamichi comes up with a solution, He notices that:
$10^{2k+1} \mod 11 = -1 (or 10)$, $10^{2k} \mod 11 = 1$,
So $X \mod 11$
$= (A_n \times 10^n + A_{n-1} \times 10^{n-1} + \ldots + A_2 \times 10^2 + A_1 \times 10^1 + A_0) \mod 11$
$= A_n \times (-1)^n + A_{n-1} \times (-1)^{n-1} + \ldots + A_2 - A_1 + A_0$
= sum_of_even_digits – sum_of_odd_digits
So he claimed that the answer is the number of numbers X in the range which satisfy the function: X mod 11 = 3. He calculate the answer in this way :
Answer = (r + 8) / 11 – (l – 1 + 8) / 11.
Rukaw heard of Hanamichi’s solution from you and he proved there is something wrong with Hanamichi’s solution. So he decided to change the test data so that Hanamichi’s solution can not pass any single test. And he asks you to do that for him.
Input
You are given a integer $T$ ($1 ≤ T ≤ 100$), which tells how many single tests the final test data has. And for the following T lines, each line contains two integers l and r, which are the original test data. ($1 ≤ l ≤ r ≤ 10^18$)
Output
Sample Input
Sample Output
题目链接
http://acm.hdu.edu.cn/showproblem.php?pid=1716
解题思路
AC代码
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