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| 1 | +Firstly, let us notice that M is small but n is very large |
| 2 | + |
| 3 | +If we have to find the answer, we have to do in terms of m, not n. |
| 4 | + |
| 5 | +Let us write the series like this |
| 6 | + |
| 7 | +(1^1 + 2^2 + 3^3 + .... + m^m) + |
| 8 | +((m + 1)^{m + 1} + (m + 2)^{m + 2} + (m + 3)^{m + 3} + ... + (2m)^{2m}) + |
| 9 | +. |
| 10 | +. |
| 11 | +. |
| 12 | + |
| 13 | +We can notice that we can write the series as a sum of M geometric progressions |
| 14 | + |
| 15 | +since (m + i)^x = i^x (mod m) |
| 16 | + |
| 17 | +----- |
| 18 | + |
| 19 | +We have to sum M series like |
| 20 | + |
| 21 | +(1^1 + 1^{m + 1} + 1^{2m + 1} + ... + ) + |
| 22 | +(2^2 + 2^{m + 2} + 2^{2m + 2} + ... + ) + |
| 23 | +. |
| 24 | +. |
| 25 | +((m - 1)^{m - 1} + (m - 1)^{2m - 1) + .... + ) |
| 26 | + |
| 27 | +The i-th geometric series has |
| 28 | + |
| 29 | +1. First term - i^i |
| 30 | +2. Difference - i^M |
| 31 | +3. Number of terms - Number of integers = i (mod M) <= N |
| 32 | + |
| 33 | +----- |
| 34 | + |
| 35 | +void solve() |
| 36 | +{ |
| 37 | + long long n, mod; |
| 38 | + cin >> n >> mod; |
| 39 | + |
| 40 | + long long sum = 0; |
| 41 | + for(int i = 1; i <= min(n, mod - 1); i++) |
| 42 | + { |
| 43 | + long long first_term = power(i, i, mod); |
| 44 | + long long multiplier = power(i, mod, mod); |
| 45 | + long long no_of_terms = (n/mod) + (n%mod >= i); |
| 46 | + |
| 47 | + sum += first_term*geometric_progression(multiplier, no_of_terms, mod); |
| 48 | + sum %= mod; |
| 49 | + } |
| 50 | + |
| 51 | + cout << sum << "\n"; |
| 52 | +} |
| 53 | + |
| 54 | +------ |
| 55 | + |
| 56 | +How to find the sum of a geometric series quickly ? |
| 57 | + |
| 58 | +We cannot use the normal expression S = a(r^n - 1)/(r - 1) |
| 59 | + |
| 60 | +since we need to calculate the inverse of (r - 1) mod M |
| 61 | + |
| 62 | +M is not guaranteed to be prime so we cannot raise the denominator to (M - 1) to get the inverse. |
| 63 | + |
| 64 | +M is large so we cannot calculate phi(M) to invoke Fermat's Little Theorem at all. |
| 65 | + |
| 66 | +Instead, we will use another beautiful method to evaluate the sum of a GP without division. |
| 67 | + |
| 68 | +----- |
| 69 | + |
| 70 | +S = 1 + r + r^2 + r^3 + .... + r^n |
| 71 | + |
| 72 | +Group (1 + r) terms |
| 73 | + |
| 74 | +S = (1 + r) + (r^2 + r^3) + (r^4 + r^5) + .... |
| 75 | + |
| 76 | +S = (1 + r) + r^2(1 + r) + r^4(1 + r) + ... |
| 77 | + |
| 78 | +S = (1 + r)(1 + r^2 + r^4 + ... ) |
| 79 | + |
| 80 | +So if the sum of a geometric series of N terms with difference r is G(r, n) |
| 81 | + |
| 82 | +G(r, n) = (1 + r)G(r^2, n/2) |
| 83 | + |
| 84 | +When there are an odd number of terms, then one power will be left out after factorizing by (1 + r) |
| 85 | + |
| 86 | +G(r, n) = (1 + r)G(r^2, n/2) + r^n |
| 87 | + |
| 88 | +We have reduced the problem of finding the sum of a term with N terms to a |
| 89 | +smaller version of itself of finding the sum of a smaller series of N/2 terms |
| 90 | + |
| 91 | +This will take O(log N) iterations to compute ! |
| 92 | + |
| 93 | +This is a very beautiful method. |
| 94 | + |
| 95 | +------ |
| 96 | + |
| 97 | +long long geometric_progression(long long multiplier, long long no_of_terms, int mod) |
| 98 | +{ |
| 99 | + //cout << "GP m = " << multiplier << " N = " << no_of_terms << "\n"; |
| 100 | + if(no_of_terms == 1) |
| 101 | + { |
| 102 | + return 1; |
| 103 | + } |
| 104 | + |
| 105 | + if(no_of_terms == 2) |
| 106 | + { |
| 107 | + return (1 + multiplier)%mod; |
| 108 | + } |
| 109 | + |
| 110 | + long long left_over_power = 0; |
| 111 | + |
| 112 | + if(no_of_terms%2 == 1) |
| 113 | + { |
| 114 | + left_over_power = power(multiplier, no_of_terms - 1, mod); |
| 115 | + } |
| 116 | + |
| 117 | + long long square_multiplier = power(multiplier, 2, mod); |
| 118 | + |
| 119 | + long long factored_gp = geometric_progression(square_multiplier, no_of_terms/2, mod); |
| 120 | + long long answer = (1 + multiplier)*factored_gp + left_over_power; |
| 121 | + |
| 122 | + |
| 123 | + return answer%mod; |
| 124 | +} |
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