Basic Topic 8 of number theory [Euclidean] [sieve prime number] [Chinese Remainder Theorem]

I have read the number theory knowledge before. Now I want to find some questions .....

Poj 1061 Portal

Here, we will give you X, Y, M, N, and L. Represents the coordinate X of frog a, y of frog B, m of frog a, n of frog B, and mod value L, find the number of encounter hops. (M-N) * X0 = (x-y) (mod L); the solution of the modulus linear equation, but pay attention to the processing, because (m-N) and (x-y) may be negative. If (m-N) is negative, a negative number is directly obtained for the two numbers. The following is a pair (x-y) + l) % L.

Then you can use the modular_linear_equation (ll a, LL B, ll N) function.

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Poj 2142 Portal

Given that Abd is three weights, AB is known weights, and now requires D, that is, M * A + N * B = d to obtain ABS (m) + for the maximum value of ABS (N), we first use the modular_linear_equation (ll a, LL B, ll N) function to obtain a set of special solutions. When M or N is the smallest, obtain the N or M of the response, and find that the minimum value of the two groups is a small value.

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Poj 2689 Portal

Detailed description

Poj 2262 Portal

The question is as follows: Any even number can be written as the sum of two prime numbers. Here, we will give you a number less than W. C will help you find the combination a + B = C, requiring a to be as small as possible, B is as big as possible.

For million data, a prime number table is used directly.

Poj 1006 portal poj 2891 Portal

For more information about the Chinese Remainder Theorem, see my blog.

HDU 1222 Portal

Question

Give M and N. A number is 0, M is added each time, and mod n is used to ask if 0 ~ N-1.

That is, to list the equation A * x = B (mod n) When B is 0, 1, 2, 3, 4 .... whether there is a solution in N-1 is actually an idea. Let's first look at the Extended Euclidean Algorithm:

// Ax = B (mod n) bool modular_linear_equation (ll a, LL B, ll N) {ll X, Y, x0, I; ll d = ex_gcd (A, N, x, Y); If (B % d! = 0) return false; X0 = x * (B/d) % N; // Special Solution for (I = 1; I <D; I ++) // The number of solutions is d printf ("% d \ n", (x0 + I * (N/D) % N); Return true ;}

In fact, when gcd (A * X, N)/B! When the value is 0, there is no solution B (0, 1, 2, 3, 4,..., n-1), that is, there is no solution when gcd (A, n) = 1.

HDU 1576 portal question: Portal

Give N, B, (n = A % 9973), and find the value of (a/B) % 9973.

N = A-A/9973*9973, A = Bx, Bx-A/9973*9973 = n. That is, Bx-9973y = n. Finally, obtain the value of X % 9973.

Extended Euclidean, bare question, if said the difficulty lies in the previous structure.