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PR003004 - Digit Sum |
For a pair of integers a and b, the digit sum of the interval [a, b] is defined as the sum of all digits occurring in all numbers between (and including) a and b. For example, the digit sum of [28, 31] can be calculated as:
2+8 + 2+9 + 3+0 + 3+1 = 28
Given the numbers a and b, calculate the digit sum of [a, b].
Input
On the first line one positive number: the number of test cases, at most 100.
After that per test case:
- one line with two space-separated integers, a and b (0 ≤ a ≤ b ≤ 1015).
Output
Per test case:
- one line with an integer: the digit sum of [a, b];
Example
Input: 3 0 10 28 31 1234 56789 Output: 46 28 1128600
Added by: | kojak_ |
Date: | 2013-11-25 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | BAPC |
hide comments
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2024-12-25 19:32:59
math |
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2024-12-13 17:47:43
pretty good starter for digit dp. |
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2022-08-25 08:05:45
easy one. :) |
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2022-08-15 18:02:03
can someone tell me if i use vector of vector as dp table it gives wa but if i use global 2d array it gives ac why? |
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2022-03-08 18:03:38
@dwij28 I had the same problem when I found that I was unintentionally converting long long int into int . Maybe you have the same problem in your code. I tried solving for 10^15 and the output was coming out to be zero, you can also try for the same value. |
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2021-08-27 17:58:40
Unsigned Long Long for the wins Thanks to everyone who commented about this and helped me save time |
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2021-07-06 11:36:58
getting wrong answer on using long long in c++ Last edit: 2021-07-06 16:37:45 |
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2021-07-06 11:35:03
@dwij28 instead of long long, use unsigned long long int |
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2021-04-29 19:44:29
Getting wrong answer. [NG]: For the sample cases you provided, you're getting correct results. Don't post any code in the comments, use forum for code review and debug. Last edit: 2021-04-30 11:51:34 |
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2020-12-24 13:15:36
easy peasy ... |