I am wanting to evaluate the expression, (an + bn + cn) % 1000000003 , in C++. I a getting overflow errors when n is very large. Can someone help me with this ? More specifically a = q + 1, b = - 2 * q and c = q - 1. I have been following the function outlined in this
Can I break (an + bn + cn) % 1000000003 into (an) % 1000000003 + (bn) % 100000003 + (cn) % 1000000003 or something similar ?
Also I cannot use anything more than unsigned long long int
You can distribute your modulo. Mathematically, this will be sound:
( ((a^n)%1000000003) + ((b^n)%100000003) + ((c^n)%1000000003) ) % 1000000003;
This will prevent you from having to compute numbers that are out of bounds, allowing you to choose larger values for n.
Proof.
Just be sure to use pow in the math.h module:
( ((pow(a, n))%1000000003)
+ ((pow(b, n))%100000003)
+ ((pow(c, n))%1000000003) ) % 1000000003;
Related
I'm stuck determining the big o notation for the below fragmented code, the given expression is part of I'm trying to figure out. I know given two plain, default for loops results in O(n^2) but the latter is entirely different. Here are the instructions.
The algorithm of
for (j = 0; j < n; j++)
{
for (k = j; k < n; k++)
{
}
}
will result in a number of iterations of given by the expression:
= n + (n-1) + (n-2) + (n-3) + ........ + (n - n)
Reduce the above series expression to an algebraic expression, without summation.
After determining the algebraic expression express the performance in Big O Notation.
You can use this method (supposedly applied by Gauss when he was a wee lad).
If you sum all the numbers twice, you have
1 + 2 + 3 + ... + n
+ n + (n-1) + (n-2) + ... + 1
—————————————————————————————————————--
(n+1) + (n+1) + (n+1) + ... + (n+1) = n(n+1)
Thus,
1 + 2 + 3 + ... + n = n(n+1)/2
and n(n+1)/2 is (n^2)/2 + n/2, so it is in O(n^2).
I currently try to get into regular expressions for school and have to work on the task to shorten this regular expression:
r = 0(e + 0 + 1)* + (e + 1)(1 + 0)* + e
with e being the empty word epsilon.
So far I got to this:
r = 0(0 + 1)* + 1(1 + 0)* + e
considering the rule
r* = (e + r)*
However, I don't really know how to continue. If it wasn't for the kleene star operators, I could use the distributive law, but that won't apply here. I can't really figure out a suitable law to continue on with this regex.
Any helpful tips?
Edit:
I think I got one step further by forming r to
r = 0(1 + 0)* + 1(1 + 0)* + e
and then being able to combine it to
r = (0 + 1)(0 + 1)* + e
Is that correct?
Also, we could then say
r = (0+1)*
which should be the final form
I'd say that your own deduction is correct except for one thing. Taking your original
r = 0(e + 0 + 1)* + (e + 1)(1 + 0)* + e
removing e which according to you is empty, leaves
r = 0(0 + 1)* + 1(1 + 0)*
or in plain words 0 followed by any number of 0 or 1 or 1 followed by any number of 1 or 0.
So the left side states that there has to be at least a 0 and the right side that there has to be a 1. That means that there must be at least a 0 or a 1. Now, your flavor of regex is one I've never seen so I don't know how to express one or more in your flavor (it's normally a +) so I'll express it in regular regex, which would be
r = [01]+
which simply means at least one0 or 1 repeated any number of times.
Regards.
There are 3 numbers: T, N, M. 1 ≤ T, M ≤ 10^9, 1 ≤ N ≤ 10^18 .
What is asked in the problem is to compute [Σ(T^i)]mod(m) where i varies from 0 to n. Obviously, O(N) or O(M) solutions wouldn't work because of 1 second time limit. How should I proceed?
As pointed out in previous answers, you may use the formula for geometric progression sum. However there is a small problem - if m is not prime, computing (T^n - 1) / (T - 1) can not be done directly - the division will not be a well-defined operations. In fact there is a solution that can handle even non prime modules and will have a complexity O(log(n) * log(n)). The approach is similar to binary exponentiation. Here is my code written in c++ for this(note that my solution uses binary exponentiation internally):
typedef long long ll;
ll binary_exponent(ll x, ll y, ll mod) {
ll res = 1;
ll p = x;
while (y) {
if (y % 2) {
res = (res * p) % mod;
}
p = (p * p) % mod;
y /= 2;
}
return res;
}
ll gp_sum(ll a, int n, ll mod) {
ll A = 1;
int num = 0;
ll res = 0;
ll degree = 1;
while (n) {
if (n & (1 << num)) {
n &= (~(1 << num));
res = (res + (A * binary_exponent(a, n, mod)) % mod) % mod;
}
A = (A + (A * binary_exponent(a, degree, mod)) % mod) % mod;
degree *= 2;
num++;
}
return res;
}
In this solution A stores consecutively the values 1, 1 + a, 1 + a + a^2 + a^3, ...1 + a + a^2 + ... a ^ (2^n - 1).
Also just like in binary exponentiation if I want to compute the sum of n degrees of a, I split n to sum of powers of two(essentially using the binary representation of n). Now having the above sequence of values for A, I choose the appropriate lengths(the ones that correspond to 1 bits of the binary representation of n) and multiply the sum by some value of a accumulating the result in res. Computing the values of A will take O(log(n)) time and for each value I may have to compute a degree of a which will result in another O(log(n)) - thus overall we have O(log(n) * log (n)).
Let's take an example - we want to compute 1 + a + a^2 .... + a ^ 10. In this case, we call gp_sum(a, 11, mod).
On the first iteration n & (1 << 0) is not zero as the first bit of 11(1011(2)) is 1. Thus I turn off this bit setting n to 10 and I accumulate in res: 0 + 1 * (a ^ (10)) = a^10. A is now a + 1.
The next second bit is also set in 10(1010(2)), so now n becomes 8 and res is a^10 + (a + 1)*(a^8)=a^10 + a^9 + a^8. A is now 1 + a + a^2 + a^3
Next bit is 0, thus res stays the same, but A will become 1 + a + a^2 + ... a^7.
On the last iteration the bit is 1 so we have:
res = a^10 + a^9 + a^8 + a^0 *(1 + a + a^2 + ... +a^7) = 1 + a .... + a ^10.
One can use an algorithm which is similar to binary exponentiation:
// Returns a pair <t^n mod m, sum of t^0..t^n mod m>,
// I assume that int is big enough to hold all values without overflowing.
pair<int, int> calc(int t, int n, int m)
if n == 0 // Base case. t^0 is always 1.
return (1 % m, 1 % m)
if n % 2 == 1
// We just compute the result for n - 1 and then add t^n.
(prevPow, prevSum) = calc(t, n - 1, m)
curPow = prevPow * t % m
curSum = (prevSum + curPow) % m
return (curPow, curSum)
// If n is even, we compute the sum for the first half.
(halfPow, halfSum) = calc(t, n / 2, m)
curPow = halfPow * halfPow % m // t^n = (t^(n/2))^2
curSum = (halfSum * halfPow + halfSum) % m
return (curPow, curSum)
The time complexity is O(log n)(the analysis is the same as for the binary exponentiation algorithm). Why is it better than a closed form formula for geometric progression? The latter involves division by (t - 1). But it is not guaranteed that there is an inverse of t - 1 mod m.
you can use this:
a^1 + a^2 + ... + a^n = a(1-a^n) / (1-a)
so, you just need to calc:
a * (1 - a^n) / (1 - a) mod M
and you can find O(logN) way to calc a^n mod M
It's a geometric series whose sum is equal to :
Hi I need to calculate (2^n + (-1)^n) % 10000007
where 1 < n < 10^9
How should I go about writing a program for it in c++?
I know this mod property
(a + b)%n = (a%n + b%n)%n but this wont help me.
Given
(a + b)%m = (a%m + b%m)%m
Then, replace both a and b with the same power of 2, and you get the recurrence:
2k+1%m = (2k%m + 2k%m)%m
You probably already figured your formula allows you to break down your problem into:
(2n + (-1)n)%P = (2n%P + (-1)n%P)%P
Then, note that (-1)k is either 1 or -1, and you should be able to calculate your problem in O(n) time.
I have the compute the sum S = (a*x + b*y + c) % N. Yes it looks like a quadratic equation but it is not because the x and y have some properties and have to be calculated using some recurrence relations. Because the sum exceeds even the limits of unsigned long long I want to know how could I compute that sum using the properties of the modulo operation, properties that allow the writing of the sum something like that(I say something because I do not remember exactly how are those properties): (a*x)%N + (b*y)%N + c%N, thus avoiding exceeding the limits of unsigned long long.
Thanks in advance for your concern! :)
a % N = x means that for some integers 0 <= x < N and m: m * N + x = a.
You can simply deduce then that if a % N = x and b % N = y then
(a + b) % N =
= (m * N + x + l * N + y) % N =
= ((m + l) * N + x + y) % N =
= (x + y) % N =
= (a % N + b % N) % N.
We know that 0 < x + y < 2N, that is why you need to keep remainder calculation. This shows that it is okay to split the summation and calculate the remainders separately and then add them, but don't forget to get the remainder for the sum.
For multiplication:
(a * b) % N =
= ((m * N + x) * (l * N + y)) % N =
= ((m * l + x * l + m * y) * N + x * y) % N =
= (x * y) % N =
= ((a % N) * (b % N)) % N.
Thus you can also do the same with products.
These properties can be simply derived in a more general setting using some abstract algebra (the remainders form a factor ring Z/nZ).
You can take the idea even further, if needed:
S = ( (a%N)*(x%N)+(b%N)*(y%N)+c%N )%N
You can apply the modulus to each term of the sum as you've suggested; but even so after summing them you must apply the modulus again to get your final result.
How about this:
int x = (7 + 7 + 7) % 10;
int y = (7 % 10 + 7 % 10 + 7 % 10) % 10;
You remember right. The equation you gave, where you %N every of the summands is correct. And that would be exactly what I use. You should also %N for every partial sum (and the total) again, as the addition results can be still greater than N. BUT be careful this works only if your size limit is at least twice as big as your N. If this is not the case, it can get really nasty.
Btw for the following %N operations of the partial sums, you dont have to perform a complete division, a check > N and if bigger just subtraction of N is enough.
Not only can you reduce all variable mod n before starting the calculation, you can write your own mod-mul to compute a*x mod n by using a shift-and-add method and reduce the result mod n at each step. That way your intermediate calculations will only require one more bit than n. Once these products are computed, you can add them pairwise and reduce mod n after each addition which will also not require more than 1 bit beyond the range of n.
There is a python implementation of modular multiplication in my answer to this question. Conversion to C should be trivial.