Poj 1006 and China Residue Theorem

The idea of poj 1006 is not very difficult. It can be transformed into a mathematical formula:

Now num is the number of days from the next day.

P, e, I, and d are set in the question!

Then we can get three sub-statements: (num + d) % 23 = p; (num + d) % 28 = e; (num + d) % 33 = I;

P, e, I, d are input, so we need to find num. For convenience, we will temporarily treat num + d as a whole! Make x = num + d;

That is, x % 23 = p; x % 28 = e; x % 33 = I; x

What should I do? This involves the so-called "China Residue Theorem" (concept google, very easy)

There is a question in Sun Tzu's computing sutra that "things do not know the number of things": "Today, things do not know the number of things. There are three, and five, and three, and seven, and two, ask things ry ?" The answer is "23 ".

-------- This is the legendary "China Residue Theorem ". In fact, the question means that n % 3 = 2, n % 5 = 3, n % 7 = 2; what is n?

So how did he solve it?

See the following:

The question involves three mass numbers: 3, 5, and 7,

Order: 5*7 * a % 3 = 1; --------------> a = 2; that is, 5*7*2 = 70;

3*7 * B % 5 = 1; --------------> B = 1; that is, 3*7*1 = 21;

3*5 * c % 7 = 1; --------------> c = 1; that is, 3*5*1 = 15;

Why do we need to set the remainder to 1: If the remainder 2 is required, multiply it by 2. If the remainder is 3, multiply it by 3!

(Because the question requires n % 3 = 2, n % 5 = 3, n % 7 = 2 ;)

So: To make n % 3 = 2, then (5*7*2) * 2% 3 = 2; (because 5*7*2% 3 = 1)

Similarly: To make n % 5 = 3, (3*7*1) * 3% 5 = 3; (because 3*7*1% 5 = 1)

Similarly: To make n % 7 = 2, (3*5*1) * 2% 7 = 2; (because 3*5*1% 7 = 1)

So what if we add (5*7*2) * 2 and (3*7*1) * 3 and (3*5*1) * 2? We know

(5*7*2) * 2 can be divisible by 5 and 7, but % 3 equals 2

(3*7*1) * 3 can be divisible by 3 and 7, but % 5 equals 3

(3*5*1) * 2 can be divisible by 3 and 5, but % 7 equals 2

Even after the sum, % 3, 5, and 7 are the same!

Then we get a number (5*7*2) * 2 + (3*7*1) * 3 + (3*5*1) * 2 = 233

But not the smallest! We still need 233% (3*5*7) = 23 solutions!

 

/*************************************** **************************************** **************************************** *******************************/

// The above is the algorithm parsing. It seems that the explanation is not very clear. Sorry ~

 

Now let's take a look at this question: x % 23 = p; x % 28 = e; x % 33 = I; x

Follow the above algorithms:

Make 33*28 * a % 23 = 1, get a = 6; 33*28*6 = 5544;

Make 23*33 * B % 28 = 1, get B = 19; 23*33*19 = 14421;
Make 23*28 * c % 33 = 1, c = 2; 23*28*2 = 1288.

Then x = 5544 * p + 14421 * e + 1288 * I

Then x-d is the number of days of the difference!

Because of the range limit, (x-d) % = 21252; and If <= 0 at this time, then (x-d) + = 21252, all of the above is to ensure that they are in the scope ~

The AC code is as follows:

[Cpp] view plaincopyprint?
/*
China Surplus theorem: from Sun Tzu's computing Sutra

*/
 
# Include <stdio. h>
 
# Deprecision MAX 21252
 
Int main ()
{
Int p, e, I, d, n, count = 0;

While (scanf ("% d", & p, & e, & I, & d )! = EOF)
{
Count ++;
If (p =-1 & e =-1 & I =-1 & d =-1)
{
Break;
}
 
N = (5544 * p + 14421 * e + 1288 * I-d) % MAX;

If (n <= 0) // range limit
{
N + = 21252;
}

Printf ("Case % d: the next triple peak occurs in % d days. \ n", count, n );
}
Return 0;
}

/*
China Surplus theorem: from Sun Tzu's computing Sutra
 
*/

# Include <stdio. h>

# Deprecision MAX 21252

Int main ()
{
Int p, e, I, d, n, count = 0;
 
While (scanf ("% d", & p, & e, & I, & d )! = EOF)
{
Count ++;
If (p =-1 & e =-1 & I =-1 & d =-1)
{
Break;
}

N = (5544 * p + 14421 * e + 1288 * I-d) % MAX;

If (n <= 0) // range limit
{
N + = 21252;
}

Printf ("Case % d: the next triple peak occurs in % d days. \ n", count, n );
}
Return 0;
}