Poj1091: flea [anti-DDoS principle]

Chinese questions are not translated

Idea: Let us assume that the flea chooses X1 first card and X2 second card... N x n cards, xn + 1 card with M reading, then the equation can be listed: A1 * X1 + A2 * X2 +... + An * xn + M * x (n + 1) = 1

As you can jump left and right, the question is to ask if the above indefinite equation has a solution? The answer and its proof can be found in any number theory book. It must be (A1, A2, A3... An, m) | 1 is A1, A2, A3... In this way, the question becomes: There are n + 1 positive integers, the maximum number of which is m, and ask how many of the sequences that meet the conditions are mutually qualitative.

Many of the mathematical combinations are difficult, but difficult. Consider how many of the largest commonalities on the back of the question is not 1? First, let's take M decomposition as a prime factor. Every factor of M can be considered as a set. The element in the set is a sequence of the maximum common divisor of this factor, the answer to this question is all the elements that are integrated in a set.

Finally, the total number of M ^ N minus the maximum number of public appointments is not 1, and the result is the number of mutual quality.

By the way, although it is part of the set theory, it is often used to prove the number theory problem. It seems to be used to derive the Euler's function formula at the earliest.

# Include <cstdio>

# Include <string>

# Include <iostream>

# Include <math. h>

# Define ll _ int64

Using namespace STD;

Llfactor [100000], H = 0, stack [100000], Top = 0, MT, NT, ret1;

Ll quickpow (ll n, ll m)

{

Ll ret = 1;

While (m)

{

If (M & 1) RET * = N;

N * = N;

M> = 1;

}

Return ret;

}

Void DFS (ll step, ll now, ll layer, ll num)

{

If (step = Layer) {ret1 + = quickpow (MT/num, NT); return ;}

For (INTI = now + 1; I <= H-layer + step + 1; I ++) DFS (Step + 1, now + 1, layer, num * factor [I]);

}

Int main ()

{

Ll N;

Scanf ("% i64d % i64d", & nt, & mt );

N = MT;

While (N & 1) = 0) {H = 1; factor [H] = 2; n >>= 1 ;}

Ll q = SQRT (N );

For (ll I = 3; I <= Q & n! = 1; I + = 2)

{

If (N % I = 0) factor [++ H] = I;

While (N % I = 0) n = N/I;

}

If (n! = 1) factor [++ H] = N;

Ll ans = 0, flag =-1;

For (ll I = 1; I <= H; I ++)

{

Flag * =-1;

Top = ret1 = 0;

DFS (0, 0, I, 1 );

Ans + = ret1 * flag;

}

Printf ("% i64d \ n", (LL) quickpow (MT, NT)-ans );

Return 0;

}

Poj1091: flea [anti-DDoS principle]