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Here 5n1+5n2

is certainly a multiple of 5, so we can write

5n1+5n2=5N where N=n1+n2

and c1+c2=C

Then from (1) we get:

(5N+C)%5

Now it is clear that the answer is C%5

. C has to be mod again because c1+c2

Its value can be greater than 5. now

((x%5)+(y%5))%5

——–(2)

=((5n1+c1)%5)+((5n2+c2)%5))%5

(5n1+c1)%5=c1

(5n2+c2)%5=c2

Then we can write 2 as:

(c1+c2)%5=C%5

Then the 1st theorem is proved. It is the same thing to mod by adding the value and mod first then add and mod again. If convenient, the number cannot be larger than any step. The same formula applies to quality.

The mod of negative numbers has to be worked a little differently. −17%5 in C

Its value shows -2. But the definition we usually use of fractions is x%m=p

If mathematically

The largest multiple of m that is x

The number obtained by subtracting the smaller number from x is p

.

ie 23%5

In this case 5×4=20 is the largest multiple of 5 which is less than 23, so 23%5=23−(5×4)=3. Note that the area of −17%5 is −20, the largest multiple of 5 that is smaller than −17, so the answer is 3.

One way to handle this case is to make the negative number a 5

Adding multiples of to make the number greater than 0, then mod. For example:

-17%5

=(-17+100)%5

=83%5

=3

It works like the proof of the formula above, just a little bit of guesswork. Always be careful in contests with negative number mod, it can be a big reason for giving wrong answer.

Now coming to the well-known big mod problem. The problem is you (ab)%m

You have to find out the value, a, b, m will be told to you, the range of all of them can be up to 2^31. 100! You can easily find the value without overflowing like finding %97, the problem is that if you want to find out 22000000000 by running the loop and multiplying one by one, you will probably finish breakfast to get the answer. If we want O(log2n)

I can do it.

notice

2100

=(250)2

And

(250)

=(225)2

Now say 250

Is it necessary to find out 226, 227 etc. or is it possible to find out up to 225 and square it? Again, to get to 225 (212)2, square it and multiply it by 2. The additional 2 times are because the number is odd. At each step the multiplication time will be mod so as not to overflow.

Writing the code using recursion is as straight as water:

#define i64 long long

i64M;

i64 F(i64 N,i64 P)

{

if(P==0) return 1;

if(P%2==0)

{

i64 ret=F(N,P/2);

return ((ret%M)*(ret%M))%M;

}

else return ((N%M)*(F(N,P-1)%M))%M;

}

In the comments section, “Hasan” linked a picture of a nice recursion-tree of a Big Mod, the picture looks like this:

Modular arithmetic is very important in mathematics as we can reduce large results to smaller ones without losing various properties of the result. Programming contests often require modular arithmetic in various problems, especially in counting and combinatorics where the results can be very large, factorials may be required.

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When dividing, formulas like multiplication and addition do not work, for this you need to know extended euclid and modular inverse.

Mod is a very costly operation for CPU. Mod takes much longer than addition, multiplication. If you use mod unnecessarily, the code may exceed the time limit, so there is no need to mod everywhere if there is no fear of overflow. After one of my codes exceeded the time limit by 3 seconds, I removed some mods and brought it down to 1.3 seconds.

Now a problem to think about. Suppose you are given a very large number (bigint) and it is 231

It is asked to mod with a smaller number. O(length_of_bigint)

How to do in complexity?

Help:

23 = (0*10+2)*10 + 3

1239=(((0*10+1)*10 + 2)*10 +3)*10+9

Problems for Practice:

http://uva.onlinejudge.org/external/3/374.html

http://uva.onlinejudge.org/external/101/10127.html

Bitwise sieve

The conventional algorithm Sieve of Eratosthene for extracting bitwise sieve prime numbers can greatly reduce the memory usage! Generating up to N primes in a simple sieve requires declaring an array of size N. Each element of the array acts as a flag that tells us whether a number is prime or composite. In Bitwise Sieve we will use bits directly instead of integers or booleans as flags.

Before reading this tutorial you need to know two things

1. Simple version of Sieve of Eratosthene, you can easily learn it by reading this post of mine.

2. Use of bitwise operator in C/C++. It can also be learned very easily from here. Read the 1st 2 parts of the tutorial very well, it will be very useful when learning Bitmask DP, BFS.

Hopefully now you know a lot about bitwise operators, you should have no problem writing simple sives. Now we will learn Bitwise Sieve.

What is the function of status or flag array in general? We can tell if a number is prime by looking at the index value of this array. I assume your status array is of integer type. Each integer has 32 bits. Why do we use so many bits to indicate empty 0 or 1? We can indicate a number with each bit at any index in the array.

When you generate primes 1-7 in an integer array, the state of your array is in binary:

status[1]=000……..00(32 zeros)

status[2]=000……..01(31 zeros, 1 1)

status[3]=000……..01(31 zeros,1 1)

status[4]=000……..00(32 zeros)

status[5]=000……..01(31 zeros,1 1)

status[6]=000……..00(32 zeros)

status[7]=000……..01(31 zeros,1 1)

31 bits in each index are unused but we can easily exploit this huge number of unused bits. We will assume:

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of status[0].

>>> Zeroth bit indicates primality of 0 (rightmost bit==0th bit)

>>> 1st bit indicates primality of 1 (rightmost bit==0th bit)

>>> 2nd bit indicates the primality of 2

>>> 3rd bit indicates the primality of 3

……………………

>>> The 31st bit indicates the primality of 31

of status[1].

>>> The zeroth bit indicates the primality of 32

>>> 1st bit indicates the primality of 33

…………..

of status[2].

>>> The zeroth bit indicates the primality of 64

>>> 1st bit indicates the primality of 65

Then generating primes from 1-7 will make your array state:

status[1]= 0000…….10101100(total 32 bits) (rightmost bit==0th bit)

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