|
| 1 | +""" |
| 2 | +Sum of digits sequence |
| 3 | +Problem 551 |
| 4 | +
|
| 5 | +Let a(0), a(1),... be an interger sequence defined by: |
| 6 | + a(0) = 1 |
| 7 | + for n >= 1, a(n) is the sum of the digits of all preceding terms |
| 8 | +
|
| 9 | +The sequence starts with 1, 1, 2, 4, 8, ... |
| 10 | +You are given a(10^6) = 31054319. |
| 11 | +
|
| 12 | +Find a(10^15) |
| 13 | +""" |
| 14 | + |
| 15 | +ks = [k for k in range(2, 20+1)] |
| 16 | +base = [10 ** k for k in range(ks[-1] + 1)] |
| 17 | +memo = {} |
| 18 | + |
| 19 | + |
| 20 | +def next_term(a_i, k, i, n): |
| 21 | + """ |
| 22 | + Calculates and updates a_i in-place to either the n-th term or the |
| 23 | + smallest term for which c > 10^k when the terms are written in the form: |
| 24 | + a(i) = b * 10^k + c |
| 25 | +
|
| 26 | + For any a(i), if digitsum(b) and c have the same value, the difference |
| 27 | + between subsequent terms will be the same until c >= 10^k. This difference |
| 28 | + is cached to greatly speed up the computation. |
| 29 | +
|
| 30 | + Arguments: |
| 31 | + a_i -- array of digits starting from the one's place that represent |
| 32 | + the i-th term in the sequence |
| 33 | + k -- k when terms are written in the from a(i) = b*10^k + c. |
| 34 | + Term are calulcated until c > 10^k or the n-th term is reached. |
| 35 | + i -- position along the sequence |
| 36 | + n -- term to caluclate up to if k is large enough |
| 37 | +
|
| 38 | + Return: a tuple of difference between ending term and starting term, and |
| 39 | + the number of terms calculated. ex. if starting term is a_0=1, and |
| 40 | + ending term is a_10=62, then (61, 9) is returned. |
| 41 | + """ |
| 42 | + # ds_b - digitsum(b) |
| 43 | + ds_b = 0 |
| 44 | + for j in range(k, len(a_i)): |
| 45 | + ds_b += a_i[j] |
| 46 | + c = 0 |
| 47 | + for j in range(min(len(a_i), k)): |
| 48 | + c += a_i[j] * base[j] |
| 49 | + |
| 50 | + diff, dn = 0, 0 |
| 51 | + max_dn = n - i |
| 52 | + |
| 53 | + sub_memo = memo.get(ds_b) |
| 54 | + |
| 55 | + if sub_memo != None: |
| 56 | + jumps = sub_memo.get(c) |
| 57 | + |
| 58 | + if jumps != None and len(jumps) > 0: |
| 59 | + # find and make the largest jump without going over |
| 60 | + max_jump = -1 |
| 61 | + for _k in range(len(jumps) - 1, -1, -1): |
| 62 | + if jumps[_k][2] <= k and jumps[_k][1] <= max_dn: |
| 63 | + max_jump = _k |
| 64 | + break |
| 65 | + |
| 66 | + if max_jump >= 0: |
| 67 | + diff, dn, _kk = jumps[max_jump] |
| 68 | + # since the difference between jumps is cached, add c |
| 69 | + new_c = diff + c |
| 70 | + for j in range(min(k, len(a_i))): |
| 71 | + new_c, a_i[j] = divmod(new_c, 10) |
| 72 | + if new_c > 0: |
| 73 | + add(a_i, k, new_c) |
| 74 | + |
| 75 | + else: |
| 76 | + sub_memo[c] = [] |
| 77 | + else: |
| 78 | + sub_memo = {c: []} |
| 79 | + memo[ds_b] = sub_memo |
| 80 | + |
| 81 | + if dn >= max_dn or c + diff >= base[k]: |
| 82 | + return diff, dn |
| 83 | + |
| 84 | + if k > ks[0]: |
| 85 | + while True: |
| 86 | + # keep doing smaller jumps |
| 87 | + _diff, terms_jumped = next_term(a_i, k - 1, i + dn, n) |
| 88 | + diff += _diff |
| 89 | + dn += terms_jumped |
| 90 | + |
| 91 | + if dn >= max_dn or c + diff >= base[k]: |
| 92 | + break |
| 93 | + else: |
| 94 | + # would be too small a jump, just compute sequential terms instead |
| 95 | + _diff, terms_jumped = compute(a_i, k, i + dn, n) |
| 96 | + diff += _diff |
| 97 | + dn += terms_jumped |
| 98 | + |
| 99 | + jumps = sub_memo[c] |
| 100 | + |
| 101 | + # keep jumps sorted by # of terms skipped |
| 102 | + j = 0 |
| 103 | + while j < len(jumps): |
| 104 | + if jumps[j][1] > dn: |
| 105 | + break |
| 106 | + j += 1 |
| 107 | + |
| 108 | + # cache the jump for this value digitsum(b) and c |
| 109 | + sub_memo[c].insert(j, (diff, dn, k)) |
| 110 | + return (diff, dn) |
| 111 | + |
| 112 | + |
| 113 | +def compute(a_i, k, i, n): |
| 114 | + """ |
| 115 | + same as next_term(a_i, k, i, n) but computes terms without memoizing results. |
| 116 | + """ |
| 117 | + if i >= n: |
| 118 | + return 0, i |
| 119 | + if k > len(a_i): |
| 120 | + a_i.extend([0 for _ in range(k - len(a_i))]) |
| 121 | + |
| 122 | + # note: a_i -> b * 10^k + c |
| 123 | + # ds_b -> digitsum(b) |
| 124 | + # ds_c -> digitsum(c) |
| 125 | + start_i = i |
| 126 | + ds_b, ds_c, diff = 0, 0, 0 |
| 127 | + for j in range(len(a_i)): |
| 128 | + if j >= k: |
| 129 | + ds_b += a_i[j] |
| 130 | + else: |
| 131 | + ds_c += a_i[j] |
| 132 | + |
| 133 | + while i < n: |
| 134 | + i += 1 |
| 135 | + addend = ds_c + ds_b |
| 136 | + diff += addend |
| 137 | + ds_c = 0 |
| 138 | + for j in range(k): |
| 139 | + s = a_i[j] + addend |
| 140 | + addend, a_i[j] = divmod(s, 10) |
| 141 | + |
| 142 | + ds_c += a_i[j] |
| 143 | + |
| 144 | + if addend > 0: |
| 145 | + break |
| 146 | + |
| 147 | + if addend > 0: |
| 148 | + add(a_i, k, addend) |
| 149 | + return diff, i - start_i |
| 150 | + |
| 151 | + |
| 152 | +def add(digits, k, addend): |
| 153 | + """ |
| 154 | + adds addend to digit array given in digits |
| 155 | + starting at index k |
| 156 | + """ |
| 157 | + for j in range(k, len(digits)): |
| 158 | + s = digits[j] + addend |
| 159 | + if s >= 10: |
| 160 | + quotient, digits[j] = divmod(s, 10) |
| 161 | + addend = addend // 10 + quotient |
| 162 | + else: |
| 163 | + digits[j] = s |
| 164 | + addend = addend // 10 |
| 165 | + |
| 166 | + if addend == 0: |
| 167 | + break |
| 168 | + |
| 169 | + while addend > 0: |
| 170 | + addend, digit = divmod(addend, 10) |
| 171 | + digits.append(digit) |
| 172 | + |
| 173 | + |
| 174 | +def solution(n): |
| 175 | + """ |
| 176 | + returns n-th term of sequence |
| 177 | +
|
| 178 | + >>> solution(10) |
| 179 | + 62 |
| 180 | +
|
| 181 | + >>> solution(10**6) |
| 182 | + 31054319 |
| 183 | +
|
| 184 | + >>> solution(10**15) |
| 185 | + 73597483551591773 |
| 186 | + """ |
| 187 | + |
| 188 | + digits = [1] |
| 189 | + i = 1 |
| 190 | + dn = 0 |
| 191 | + while True: |
| 192 | + diff, terms_jumped = next_term(digits, 20, i + dn, n) |
| 193 | + dn += terms_jumped |
| 194 | + if dn == n - i: |
| 195 | + break |
| 196 | + |
| 197 | + a_n = 0 |
| 198 | + for j in range(len(digits)): |
| 199 | + a_n += digits[j] * 10 ** j |
| 200 | + return a_n |
| 201 | + |
| 202 | + |
| 203 | +if __name__ == "__main__": |
| 204 | + print(solution(10 ** 15)) |
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