poj1006 - 爱码网

数论跪了三天。。

这个题不难得到(n+d)%23=p; (n+d)%28=e; (n+d)%33=i

如何求解？

首先介绍一个所谓“逆”的概念。

给定整数a，有（a，m）=1，称ax=1(mod m)的一个解叫做a模m的逆。

下面给出求逆的程序。

#include <iostream>
#include <math.h>

using namespace std;

typedef long long LL;

void gcd(LL a, LL b, LL &d, LL &x, LL &y)
{
    if(!b)
    {
        d = a, x = 1, y = 0;
    }
    else
    {
        gcd(b, a %b, d, y, x);
        y -= x * (a/b);
    }
}
LL inv(LL a, LL n)
{
    LL d, x, y;
    gcd(a,n,d,x,y);
    return d == 1 ? (x + n) % n : -1;
}
int main()
{
    LL a, m;
    while(cin>> a>>m)
    {
        cout << inv(a, m) <<endl;
    }
}

比如9模31的逆是7。可输入9 31，得出7。

“中国剩余定理”

http://en.wikipedia.org/wiki/Chinese_remainder_theorem

百度百科介绍的就是一坨屎，看都别看！

言归正传，如何解这个方程：

(n+d)%23=p;(n+d)%28=e;(n+d)%33=i;

poj1006

大家看到了，都是让它的余数变成1（为了求逆）。“大衍求一术”。

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Finding the solution with basic algebra and modular arithmetic

For example, consider the problem of finding an integer x such that

$poj1006$

A brute-force approach converts these congruences into sets and writes the elements out to the product of 3×4×5 = 60 (the solutions modulo 60 for each congruence):

x ∈ {2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, …}

x ∈ {3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, …}

x ∈ {1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, …}

To find an x that satisfies all three congruences, intersect the three sets to get:

x ∈ {11, …}

Which can be expressed as

$poj1006$

Another way to find a solution is with basic algebra, modular arithmetic, and stepwise substitution.

We start by translating these congruences into equations for some t, s, and u:

Equation 1: $poj1006$
Equation 2: $poj1006$
Equation 3: $poj1006$

Start by substituting the x from equation 1 into congruence 2:

$poj1006$

meaning that $poj1006$ for some integer s.

Plug t into equation 1:

$poj1006$

Plug this x into congruence 3:

$poj1006$

Casting out fives, we get

$poj1006$

meaning that

$poj1006$

for some integer u.

Finally,

$poj1006$

So, we have solutions 11, 71, 131, 191, …

Notice that 60 = lcm(3,4,5). If the moduli are pairwise coprime (as they are in this example), the solutions will be congruent modulo their product.

A constructive algorithm to find the solution

The following algorithm only applies if the $poj1006$ 's are pairwise coprime. (For simultaneous congruences when the moduli are not pairwise coprime, themethod of successive substitution can often yield solutions.)

Suppose, as above, that a solution is required for the system of congruences:

$poj1006$

Again, to begin, the product $poj1006$ is defined. Then a solution x can be found as follows.

For each i the integers $poj1006$ and $poj1006$ are coprime. Using the extended Euclidean algorithm we can find integers $poj1006$ and $poj1006$ such that $poj1006$ . Then, choosing the label $poj1006$ , the above expression becomes:

$poj1006$

Consider $poj1006$ . The above equation guarantees that its remainder, when divided by $poj1006$ , must be 1. On the other hand, since it is formed as $poj1006$ , the presence of N guarantees a remainder of zero when divided by any $poj1006$ when $poj1006$ .

$poj1006$

Because of this, and the multiplication rules allowed in congruences, one solution to the system of simultaneous congruences is:

$poj1006$

For example, consider the problem of finding an integer x such that

$poj1006$

Using the extended Euclidean algorithm, for x modulo 3 and 20 [4×5], we find (−13) × 3 + 2 × 20 = 1; i.e., e₁ = 40. For x modulo 4 and 15 [3×5], we get (−11) × 4 + 3 × 15 = 1, i.e. e₂ = 45. Finally, for x modulo 5 and 12 [3×4], we get 5 × 5 + (−2) × 12 = 1, i.e. e₃ = −24. A solution x is therefore 2 × 40 + 3 × 45 + 1 × (−24) = 191. All other solutions are congruent to 191 modulo 60, [3 × 4 × 5 = 60], which means they are all congruent to 11 modulo 60.

Note: There are multiple implementations of the extended Euclidean algorithm which will yield different sets of $poj1006$ , $poj1006$ , and $poj1006$ . These sets however will produce the same solution; i.e., (−20)2 + (−15)3 + (−24)1 = −109 = 11 modulo 60.

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那么，程序很简单

注意一点，如果直接对21252取余，再减去d，可能会出现负数的情况。

#include<iostream>

using namespace std;

int main()
{
	int p, e, i, d;
	int time = 1;
	while (cin >> p >> e >> i >> d)
	{
		if (p == -1 && e == -1 && i == -1 && d == -1)
			break;
		int n = (5544 * p + 14421 * e + 1288 * i - d + 21252) % 21252;
		if (n == 0)
			n = 21252;
		cout << "Case " << time++ << ": the next triple peak occurs in " << n << " days." << endl;
	}
}

请试着解下列同余方程组作为练习

poj1006