#include <bits/stdc++.h> using namespace std; struct number { long long int val; long long int pref; int index; //in days int split[3]; vector<long long int> count; }; int multiplyDigits(long long int a) { long long int result = 1; while(a != 0) { result *= (a%10); a /= 10; } return (int)result; } void output(int t, vector<number> &days) { sort(days.begin(), days.end(), [](number &a, number &b) {return a.index < b.index;}); for(int j=0; j<t; j++) { number &n = days[j]; for(auto i : n.count) { cout << i << " "; } cout << "\n"; } } int main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); int t; //days cin >> t; vector<number> days(t+1); days[t].val = LONG_LONG_MAX; days[t].pref = LONG_LONG_MAX; days[t].split[0] = INT_MAX; days[t].index = INT_MAX; int mil = 1000000; for(int i=0; i<t; i++) { cin >> days[i].val; days[i].index = i; long long int temp = days[i].val; days[i].split[2] = (int)(temp%mil); temp /= mil; days[i].pref = temp; days[i].split[1] = (int)(temp%mil); days[i].split[0] = (int)(temp/mil); days[i].count.assign(10, 0); } sort(days.begin(), days.end(), [](number &a, number &b) {return a.val < b.val;}); vector<int> roots(mil); for(int i=1; i<mil; i++) { roots[i] = multiplyDigits(i); } vector<long long int> counter(10, 0); int index = 0; //which number are we looking for for(int i=1; i<mil; i++) { int val = i; while(val > 9) { val = roots[val]; } counter[val]++; while(days[index].val == i) { days[index].count = counter; index++; } } if(index == t) { output(t, days); return 0; } //else -> something greater than a mil for(int j=1; j<mil; j++) { int cur = roots[j]; if(cur == 0 && days[index].pref != j) { counter[0] += mil; continue; } //else -> need to check for(int i=0; i<mil; i++) { long long int val; if(i < 100000) { val = 0; } else { val = cur*roots[i]; while(val >= mil) val = multiplyDigits(val); while(val > 9) { val = roots[val]; } } counter[val]++; while(i == days[index].split[2] && j == days[index].pref) { days[index].count = counter; index++; if(index == t) { output(t, days); return 0; } } } } int rest = mil*mil; //brute from now on, too slow anyway for(;;rest++) { counter[multiplyDigits(rest)]++; while(rest == days[index].val) { days[index].count = counter; index++; if(index == t) { output(t, days); return 0; } } } return 0; } /* https://math.stackexchange.com/questions/4731838/patterns-after-recursive-multiplication-of-the-digits-of-a-number-until-one-digi recursive digit product A031347 - multiplicative digital root */ /* If at the beginning (in mil) 0: 902609 1: 6 2: 16673 3: 21 4: 2345 5: 2073 6: 43538 7: 21 8: 32658 9: 56 Otherwise 0: 920207 1: 1 2: 13460 3: 6 4: 1451 5: 1466 6: 36695 7: 6 8: 26687 9: 21 */
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 | #include <bits/stdc++.h> using namespace std; struct number { long long int val; long long int pref; int index; //in days int split[3]; vector<long long int> count; }; int multiplyDigits(long long int a) { long long int result = 1; while(a != 0) { result *= (a%10); a /= 10; } return (int)result; } void output(int t, vector<number> &days) { sort(days.begin(), days.end(), [](number &a, number &b) {return a.index < b.index;}); for(int j=0; j<t; j++) { number &n = days[j]; for(auto i : n.count) { cout << i << " "; } cout << "\n"; } } int main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); int t; //days cin >> t; vector<number> days(t+1); days[t].val = LONG_LONG_MAX; days[t].pref = LONG_LONG_MAX; days[t].split[0] = INT_MAX; days[t].index = INT_MAX; int mil = 1000000; for(int i=0; i<t; i++) { cin >> days[i].val; days[i].index = i; long long int temp = days[i].val; days[i].split[2] = (int)(temp%mil); temp /= mil; days[i].pref = temp; days[i].split[1] = (int)(temp%mil); days[i].split[0] = (int)(temp/mil); days[i].count.assign(10, 0); } sort(days.begin(), days.end(), [](number &a, number &b) {return a.val < b.val;}); vector<int> roots(mil); for(int i=1; i<mil; i++) { roots[i] = multiplyDigits(i); } vector<long long int> counter(10, 0); int index = 0; //which number are we looking for for(int i=1; i<mil; i++) { int val = i; while(val > 9) { val = roots[val]; } counter[val]++; while(days[index].val == i) { days[index].count = counter; index++; } } if(index == t) { output(t, days); return 0; } //else -> something greater than a mil for(int j=1; j<mil; j++) { int cur = roots[j]; if(cur == 0 && days[index].pref != j) { counter[0] += mil; continue; } //else -> need to check for(int i=0; i<mil; i++) { long long int val; if(i < 100000) { val = 0; } else { val = cur*roots[i]; while(val >= mil) val = multiplyDigits(val); while(val > 9) { val = roots[val]; } } counter[val]++; while(i == days[index].split[2] && j == days[index].pref) { days[index].count = counter; index++; if(index == t) { output(t, days); return 0; } } } } int rest = mil*mil; //brute from now on, too slow anyway for(;;rest++) { counter[multiplyDigits(rest)]++; while(rest == days[index].val) { days[index].count = counter; index++; if(index == t) { output(t, days); return 0; } } } return 0; } /* https://math.stackexchange.com/questions/4731838/patterns-after-recursive-multiplication-of-the-digits-of-a-number-until-one-digi recursive digit product A031347 - multiplicative digital root */ /* If at the beginning (in mil) 0: 902609 1: 6 2: 16673 3: 21 4: 2345 5: 2073 6: 43538 7: 21 8: 32658 9: 56 Otherwise 0: 920207 1: 1 2: 13460 3: 6 4: 1451 5: 1466 6: 36695 7: 6 8: 26687 9: 21 */ |