https://www.spoj.com/problems/CPCRC1C/
https://www.geeksforgeeks.org/digit-dp-introduction/
The basic idea for state relation is very simple. We formulate the dp in top-down fashion.
Let’s say we are at the msd having index idx. So initially sum will be 0.
https://www.quora.com/How-do-you-solve-this-problem-using-DP-in-digits
Majid is a 3rd-grade elementary student and quite well in mathematics. Once, Majid's teacher asked him to calculate the sum of numbers 1 through n.
Majid quickly answered, and his teacher made him another challenge. He asked Majid to calculate the sum of the digits of numbers 1 through n.
Majid did finally find the solution. Now it is your turn, can you find a solution?
Input
Two space-separated integers 0 <= a <= b <= 109.
Program terminates if a and b are -1.
Output
The sum of the digits of numbers a through b.
Example
Input: 1 10 100 777 -1 -1 Output: 46 8655
Now we see that if we have calculated the answer for state having n-1 digits, i.e., tn-1 tn-2… t2 t1 and we need to calculate answer for state having n digitdtn tn-1 tn-2 … t2 t1. So, clearly, we can use the result of the previous state instead of re-calculating it. Hence, it follows the overlapping property.
Let’s think for a state for this DP
Our DP state will be dp(idx, tight, sum)
1) idx
- It tells about the index value from right in the given integer
2) tight
- This will tell if the current digits range is restricted or not. If the current digit’s
range is not restricted then it will span from 0 to 9 (inclusively) else it will span
from 0 to digit[idx] (inclusively).Example: consider our limiting integer to be 3245 and we need to calculate G(3245)
index : 4 3 2 1
digits : 3 2 4 5
Unrestricted range:
Now suppose the integer generated till now is : 3 1 * * ( * is empty place, where digits are to be inserted to form the integer).
Now suppose the integer generated till now is : 3 1 * * ( * is empty place, where digits are to be inserted to form the integer).
index : 4 3 2 1 digits : 3 2 4 5 generated integer: 3 1 _ _
here, we see that index 2 has unrestricted range. Now index 2 can have digits from range 0 to 9(inclusively).
For unrestricted range tight = 0
For unrestricted range tight = 0
Restricted range:
Now suppose the integer generated till now is : 3 2 * * ( ‘*’ is an empty place, where digits are to be inserted to form the integer).
Now suppose the integer generated till now is : 3 2 * * ( ‘*’ is an empty place, where digits are to be inserted to form the integer).
index : 4 3 2 1 digits : 3 2 4 5 generated integer: 3 2 _ _
here, we see that index 2 has a restricted range. Now index 2 can only have digits from range 0 to 4 (inclusively)
For unrestricted range tight = 1
For unrestricted range tight = 1
3) sum
- This parameter will store the sum of digits in the generated integer from msd to idx.
- Max value for this parameter sum can be 9*18 = 162, considering 18 digits in the integer
State Relation
The basic idea for state relation is very simple. We formulate the dp in top-down fashion.
Let’s say we are at the msd having index idx. So initially sum will be 0.
Therefore, we will fill the digit at index by the digits in its range. Let’s say its range is from 0 to k (k<=9, depending on the tight value) and fetch the answer from the next state having index = idx-1 and sum = previous sum + digit chosen.
int ans = 0;
for (int i=0; i<=k; i++) {
ans += state(idx-1, newTight, sum+i)
}
state(idx,tight,sum) = ans;
How to calculate the newTight value?
The new tight value from a state depends on its previous state. If tight value form the previous state is 1 and the digit at idx chosen is digit[idx](i.e the digit at idx in limiting integer) , then only our new tight will be 1 as it only then tells that the number formed till now is prefix of the limiting integer.
The new tight value from a state depends on its previous state. If tight value form the previous state is 1 and the digit at idx chosen is digit[idx](i.e the digit at idx in limiting integer) , then only our new tight will be 1 as it only then tells that the number formed till now is prefix of the limiting integer.
// digitTaken is the digit chosen // digit[idx] is the digit in the limiting // integer at index idx from right // previouTight is the tight value form previous // state newTight = previousTight & (digitTake == digit[idx])
There are total idx*sum*tight states and we are performing 0 to 9 iterations to visit every state. Therefore, The Time Complexity will be O(10*idx*sum*tight). Here, we observe that tight = 2 and idx can be max 18 for 64 bit unsigned integer and moreover, the sum will be max 9*18 ~ 200. So, overall we have 10*18*200*2 ~ 10^5 iterations which can be easily executed in 0.01 seconds.
