implementation *1000

Please click on ads to support us..

Python Code:



d = [6,2,5,5,4,5,6,3,7,6]
a,b = map(int,input().split())
num = str(list(range(a,b+1)))
ans = 0
for i in range(len(d)):
    ans+=d[i]*num.count(str(i))
print(ans)

C++ Code:

#include<bits/stdc++.h>
using namespace std;
#define ll long long

int findnum(int n){
    int digit[10] = {6, 2, 5, 5, 4, 5, 6, 3, 7, 6};
    int sum = 0;
    while(n>0){
        sum = sum + digit[n % 10];
        n = n / 10;
    }

    return sum;
}


int main(){
    int a ,b;
    // cout b;
    cin >> a >> b;
    ll total = 0;
    for (int i = a; i <= b;i++){
        total = total + findnum(i);
    }

    cout << total;
}


Comments

Submit
0 Comments
More Questions

478A - Initial Bet
981A - Antipalindrome
365A - Good Number
1204B - Mislove Has Lost an Array
1409D - Decrease the Sum of Digits
1476E - Pattern Matching
1107A - Digits Sequence Dividing
1348A - Phoenix and Balance
1343B - Balanced Array
1186A - Vus the Cossack and a Contest
1494A - ABC String
1606A - AB Balance
1658C - Shinju and the Lost Permutation
1547C - Pair Programming
550A - Two Substrings
797B - Odd sum
1093A - Dice Rolling
1360B - Honest Coach
1399C - Boats Competition
1609C - Complex Market Analysis
1657E - Star MST
1143B - Nirvana
1285A - Mezo Playing Zoma
919B - Perfect Number
894A - QAQ
1551A - Polycarp and Coins
313A - Ilya and Bank Account
1469A - Regular Bracket Sequence
919C - Seat Arrangements
1634A - Reverse and Concatenate