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July 11, 2019 10:50
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| // DigitDP variation for calculating the sum of digits - CPCRC1C on SPOJ | |
| // Code inspired by GeeksforGeeks - coded by adist98 | |
| // Given two integers a and b. The task is to print | |
| // sum of all the digits appearing in the | |
| // integers between a and b | |
| #include "bits/stdc++.h" | |
| using namespace std; | |
| // Memoization for the state results | |
| long long int dp[20][180][2]; | |
| // Stores the digits in x in a vector digit | |
| long long int getDigits(long long int x, vector <long long int> &digit) | |
| { | |
| while (x) | |
| { | |
| digit.push_back(x%10); | |
| x /= 10; | |
| } | |
| } | |
| // Return sum of digits from 1 to integer in | |
| // digit vector | |
| long long int digitSum(int idx, long long int sum, int tight, | |
| vector <long long int> &digit) | |
| { | |
| // base case | |
| if (idx == -1) | |
| return sum; | |
| // checking if already calculated this state | |
| if (dp[idx][sum][tight] != -1 && tight != 1) | |
| return dp[idx][sum][tight]; | |
| long long ret = 0; | |
| // calculating range value | |
| int k = (tight)? digit[idx] : 9; | |
| for (int i = 0; i <= k ; i++) | |
| { | |
| // caclulating newTight value for next state | |
| int newTight = (digit[idx] == i)? tight : 0; | |
| // fetching answer from next state | |
| ret += digitSum(idx-1, sum+i, newTight, digit); | |
| } | |
| if (!tight) | |
| dp[idx][sum][tight] = ret; | |
| return ret; | |
| } | |
| // Returns sum of digits in numbers from a to b. | |
| long long int rangeDigitSum(long long int a, long long int b) | |
| { | |
| // initializing dp with -1 | |
| memset(dp, -1, sizeof(dp)); | |
| // storing digits of a-1 in digit vector | |
| vector<long long int> digitA; | |
| getDigits(a-1, digitA); | |
| // Finding sum of digits from 1 to "a-1" which is passed | |
| // as digitA. | |
| long long int ans1 = digitSum(digitA.size()-1, 0, 1, digitA); | |
| // Storing digits of b in digit vector | |
| vector<long long int> digitB; | |
| getDigits(b, digitB); | |
| // Finding sum of digits from 1 to "b" which is passed | |
| // as digitB. | |
| long long int ans2 = digitSum(digitB.size()-1, 0, 1, digitB); | |
| return (ans2 - ans1); | |
| } | |
| // driver function to call above function | |
| int main() | |
| { | |
| int a,b; | |
| while(true){ | |
| cin >> a >> b; | |
| if(a == -1 && b == -1){ | |
| break; | |
| } | |
| cout << rangeDigitSum(a, b) << endl; | |
| } | |
| return 0; | |
| } | |
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