Sicily 7969. Digit Sum

7969. Digit Sum

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

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].

Sample Input

30 1028 311234 56789

Sample Output

46281128600

Problem Source

2013年每周一赛第五场暨校赛模拟赛III/BAPC 2012

// Problem#: 7969// Submission#: 3593648// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/// All Copyright reserved by Informatic Lab of Sun Yat-sen University// Solution to Digit Sum// Author: Thomas Beuman// Time complexity: O(log b)// Memory: O(1)// @EXPECTED_RESULTS@: CORRECT/*Solution outline:Compute the digitsum from 0 to b and from 0 to a-1, and subtract the twoDetermine the digitsum from 0 to n-1 recursively:Example:digitsum(125) = digitsum(120) + sumofdigits(120) + sumofdigits(121) + ... + sumofdigits(124)  digitsum(120) = 10 * digitsum(12) + (0+1+2+3+4+5+6+7+8+9) * 12  sumofdigits(120) + sumofdigits(121) + ... + sumofdigits(124) = 5 * sumofdigits(12) + (0+1+2+3+4)*/#include <cstdio>using namespace std;typedef long long i64;// Returns the sum of the digits of ni64 sumofdigits (i64 n){ return n ? n%10 + sumofdigits(n/10) : 0;}// Returns the sum of all digits from 0 to n-1i64 digitsum (i64 n){ return n < 10 ? n*(n-1)/2                : digitsum(n%10) + 45 * (n/10) // Last digit                  + (n%10) * sumofdigits(n/10) + 10 * digitsum(n/10); // Other digits}int main(){ int cases, casenr;  i64 a, b;  scanf("%d", &cases);  for (casenr = 1; casenr <= cases; casenr++)  { scanf("%lld %lld", &a, &b);    printf("%lld\n", digitsum(b+1) - digitsum(a));  }  return 0;}