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| 1 | +/* |
| 2 | +We define super digit of an integer using the following rules: |
| 3 | +
|
| 4 | +Given an integer, we need to find the super digit of the integer. |
| 5 | +
|
| 6 | +If has only digit, then its super digit is . |
| 7 | +Otherwise, the super digit of is equal to the super digit of the sum of the digits of . |
| 8 | +For example, the super digit of will be calculated as: |
| 9 | +
|
| 10 | + super_digit(9875) 9+8+7+5 = 29 |
| 11 | + super_digit(29) 2 + 9 = 11 |
| 12 | + super_digit(11) 1 + 1 = 2 |
| 13 | + super_digit(2) = 2 |
| 14 | +You are given two numbers and . The number is created by concatenating the string times. Continuing the above example where , assume your value . Your initial (spaces added for clarity). |
| 15 | +
|
| 16 | + superDigit(p) = superDigit(9875987598759875) |
| 17 | + 5+7+8+9+5+7+8+9+5+7+8+9+5+7+8+9 = 116 |
| 18 | + superDigit(p) = superDigit(116) |
| 19 | + 1+1+6 = 8 |
| 20 | + superDigit(p) = superDigit(8) |
| 21 | +All of the digits of sum to . The digits of sum to . is only one digit, so it's the super digit. |
| 22 | +
|
| 23 | +Function Description |
| 24 | +
|
| 25 | +Complete the function superDigit in the editor below. It must return the calculated super digit as an integer. |
| 26 | +
|
| 27 | +superDigit has the following parameter(s): |
| 28 | +
|
| 29 | +n: a string representation of an integer |
| 30 | +k: an integer, the times to concatenate to make |
| 31 | +Input Format |
| 32 | +
|
| 33 | +The first line contains two space separated integers, and . |
| 34 | +
|
| 35 | +Constraints |
| 36 | +
|
| 37 | +Output Format |
| 38 | +
|
| 39 | +Return the super digit of , where is created as described above. |
| 40 | +
|
| 41 | +Sample Input 0 |
| 42 | +
|
| 43 | +148 3 |
| 44 | +Sample Output 0 |
| 45 | +
|
| 46 | +3 |
| 47 | +Explanation 0 |
| 48 | +
|
| 49 | +Here and , so . |
| 50 | +
|
| 51 | +super_digit(P) = super_digit(148148148) |
| 52 | + = super_digit(1+4+8+1+4+8+1+4+8) |
| 53 | + = super_digit(39) |
| 54 | + = super_digit(3+9) |
| 55 | + = super_digit(12) |
| 56 | + = super_digit(1+2) |
| 57 | + = super_digit(3) |
| 58 | + = 3. |
| 59 | +Sample Input 1 |
| 60 | +
|
| 61 | +9875 4 |
| 62 | +Sample Output 1 |
| 63 | +
|
| 64 | +8 |
| 65 | +Sample Input 2 |
| 66 | +
|
| 67 | +123 3 |
| 68 | +Sample Output 2 |
| 69 | +
|
| 70 | +9 |
| 71 | +Explanation 2 |
| 72 | +
|
| 73 | +Here and , so . |
| 74 | +
|
| 75 | +super_digit(P) = super_digit(123123123) |
| 76 | + = super_digit(1+2+3+1+2+3+1+2+3) |
| 77 | + = super_digit(18) |
| 78 | + = super_digit(1+8) |
| 79 | + = super_digit(9) |
| 80 | + = 9 |
| 81 | +*/ |
| 82 | + |
| 83 | +#include <bits/stdc++.h> |
| 84 | + |
| 85 | +using namespace std; |
| 86 | + |
| 87 | +vector<string> split_string(string); |
| 88 | + |
| 89 | +// Complete the superDigit function below. |
| 90 | +int superDigit(string n, int k) { |
| 91 | + |
| 92 | + if(n.size() == 1 && k == 1) return n[0]-'0'; |
| 93 | + |
| 94 | + long long sum = 0; |
| 95 | + for(char c:n) |
| 96 | + sum+=(c-'0'); |
| 97 | + |
| 98 | + sum*=k; |
| 99 | + |
| 100 | + while(sum > 10){ |
| 101 | + long long s = 0; |
| 102 | + while(sum > 0){ |
| 103 | + s+=sum%10; |
| 104 | + sum/=10; |
| 105 | + } |
| 106 | + sum = s; |
| 107 | + } |
| 108 | + |
| 109 | + return (int)sum; |
| 110 | +} |
| 111 | + |
| 112 | +int main() |
| 113 | +{ |
| 114 | + ofstream fout(getenv("OUTPUT_PATH")); |
| 115 | + |
| 116 | + string nk_temp; |
| 117 | + getline(cin, nk_temp); |
| 118 | + |
| 119 | + vector<string> nk = split_string(nk_temp); |
| 120 | + |
| 121 | + string n = nk[0]; |
| 122 | + |
| 123 | + int k = stoi(nk[1]); |
| 124 | + |
| 125 | + int result = superDigit(n, k); |
| 126 | + |
| 127 | + fout << result << "\n"; |
| 128 | + |
| 129 | + fout.close(); |
| 130 | + |
| 131 | + return 0; |
| 132 | +} |
| 133 | + |
| 134 | +vector<string> split_string(string input_string) { |
| 135 | + string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) { |
| 136 | + return x == y and x == ' '; |
| 137 | + }); |
| 138 | + |
| 139 | + input_string.erase(new_end, input_string.end()); |
| 140 | + |
| 141 | + while (input_string[input_string.length() - 1] == ' ') { |
| 142 | + input_string.pop_back(); |
| 143 | + } |
| 144 | + |
| 145 | + vector<string> splits; |
| 146 | + char delimiter = ' '; |
| 147 | + |
| 148 | + size_t i = 0; |
| 149 | + size_t pos = input_string.find(delimiter); |
| 150 | + |
| 151 | + while (pos != string::npos) { |
| 152 | + splits.push_back(input_string.substr(i, pos - i)); |
| 153 | + |
| 154 | + i = pos + 1; |
| 155 | + pos = input_string.find(delimiter, i); |
| 156 | + } |
| 157 | + |
| 158 | + splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1)); |
| 159 | + |
| 160 | + return splits; |
| 161 | +} |
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