remainder theorem exercises

Explanation of Chinese Remainder Theorem, explanation of Remainder Theorem China Remainder Theorem There is a question in Sun Tzu's Computing Theory: "Today there are things that don't know the number of things. There are two thre

The idea of POJ 1006 is not very difficult and can be transformed into mathematical formula:Now num is the number of days from the next same day to the beginningP,e,i,d title in the set!Then you can get three formulas: (num + d)% = = p, (num + d)% = = e; (num + d)% = = i;P,e,i,d is our input, then we need to find Num can, for convenience, we will num+d temporarily as a whole! make x = num + D;namely: x% = = p; x% = = e; x% = = i; xWhat to do? This involves the so-called " Chinese

1951: [Sdoi2010] Ancient pig wen Seeking G^sigma{c (N, i), I | N} The value of mod m, where m = 999911659.M is a prime number, so according to Fermat theorem g^ (M-1) ≡1 (mod M).Then G^sigma{c (N, i), I | n}≡g^ (Sigma{c (N, i), I | n} mod (M-1)) ≡g^sigma{c (n, i) mod (M-1), I | N} (mod M). So we only need C (N, K) mod (M-1) and then accumulate.Note: M-1 = 2 * 3 * 4679 * 35617.So we can getP≡A1 (mod 2)P≡A2 (mod 3)P≡A3 (mod 4679)P≡A4 (mod 35617)After t

-1. This is obvious, because it is equivalent to P divisible, that is, P divisible (x + 1) (x-1 ). Since P is a prime number, it is only possible that the X-1 can be divisible by P (at this time x = 1) or x + 1 can be divisible by P (at this time x = PM ). The following example shows how the theorem is applied to the Fermat test. As mentioned above, 341 can pass a 2-based Fermat test, because 2 ^ 340 mod 341 = 1. If 341 is a prime number, 2 ^ 170mod 3

, when k equals a certain value, the number of characters of the pig in the face is n/k.。 However, it is also quite a lot to keep n/k from n characters. Ipig predicts that if all the possible k's cases add up to p, then the cost of studying ancient writings will be the P-th side of G. Now he wants to know what the cost of studying ancient writings in the Pig Kingdom is. Since Ipig thinks this number can be astronomical, you just need to tell him that the answer is divided by the

=sqrt(n) +1; iif(! (n%i)){ for(intj=0;j4; j + +) num[j]= (Num[j]+lucas (N,i,prime[j]))%prime[j];if(I*i-N) for(intj=0;j4; j + +) num[j]= (Num[j]+lucas (N,n/i,prime[j]))%prime[j]; } ll Mul=1, ans=0; for(intI=0;i4; i++) Mul*=Prime[i]; for(LL i=0,t;i4; i++) {EXGCD (mul/prime[i],prime[i],inver[i],t); Inver[i]*=Mul/prime[i]; } for(intI=0;i4; i++) ans= (Ans+num[i]*inver[i])%(mod-1);return(Ans+ (mod-1))%(mod-1);}intMain () {intN,g; scanf"%d%d", n,g);if(mod==g) {printf("0");return 0; } g%=MoDprintf("%lld

much, because the history is too long, has not been verified. Ipig think that as long as the literature, each can divide n k is possible. He intends to take into account all possible K. Obviously when K is equal to a certain value, the number of characters of the pig in this direction is N/k. however, It is also quite a lot to keep n/k from N characters. Ipig predicts that if all the possible K's cases add up to p, then the cost of studying ancient writings will be the p-th side of G.Now he wan

Link: poj 1845 Evaluate the sum of all the factors of a ^ B and the remainder of 9901 For example, the factors of 2 ^ 3 = are 1, 2, 4, 8, and the sum is 15. 15 is the result after the remainder is obtained. There are three main application theorems: (1) unique factorization theorem of integers: Any positive integer has only one way to write the product expression

Big numberTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 5930 Accepted Submission (s): 4146Problem Descriptionas We know, Big number is always troublesome. But it's really important in our ACM. And today, your task is to write a program to calculate a mod B.The problem easier, I promise that B'll be smaller than 100000.Is it too hard? No, I work it out in the minutes, and my program contains less than lines.Inputthe input contains several test

Examples of Chinese Remainder Theorem algorithm implemented by Python and python theorem Algorithm This example describes the Chinese Remainder Theorem algorithm implemented by Python. We will share this with you for your reference. The details are as follows: Chinese

"Chinese remainder theorem" is a famous arithmetic work in the period of 5-6 century in China's northern and Southern dynasties, a "matter unknown number" in the book of grandchildren, a solution to the problem: Today there is no know its number, 33 of the remaining two,55 of the remaining three, 77 of the remaining two. Asking for geometry? Answer: 23.According to the above we can get a set of formulas:X≡2

Explanation of extended Chinese Remainder Theorem and explanation of Chinese TheoremPreface Before reading this article, we recommend that you first learn the Chinese Remainder Theorem. In fact, it doesn't matter if you don't learn it. After all, there is no relationship between the two.Extended CRT We know that the Ch

combination on a line.Sample Input19 5 23 5Sample OUTPUT6Source2015 ACM/ICPC Asia Regional Changchun OnlineRecommendhujie | We have a carefully selected several similar problems for you:5842 5841 5840 5839 5838 Analysis: According to Lucas solution each I:C (n,m)%pi, and then according to the The state surplus theorem integrates these results. will be able to get answers. Note that the Chinese remainder

Today, I am bored occasionally. When I see something called the "China Surplus Theorem", I think it is fun to write a summary of the question first. Suppose that a number (1) is divided by 3 and 2, (2) divide by 5 and 4, (3) divide by 7 and 6, and obtain the minimum integer lcm (5, 7) that meets the conditions) 35 35*2 = 70 just divided by 3 remainder 1 lcm (3, 7) is 21 21 21*1 = 21 just divided by 5

Introduction to China Residue Theorem In Sun Tzu's computing Sutra, there is a problem: "I don't know the number of things today, but the number of three is two (divided by more than three 2 ), three of the five remaining values (divided by more than 5 3), two of the seven remaining values (divided by more than 7 2), and asked ry?" This is called the "Sun Tzu's question ", The Ming Dynasty mathematician, Cheng Damai, hinted at the solution of this que

The original topic see HDU 5768 In the [L,r] range is a multiple of 7 and does not satisfy the number of any given congruence. If the range is [1,100], does not meet the modulus of 3 to 2 or more than 3 of the modulus of 5 is 7,21,42,49,70,84,91, so the answer is 7.Where the divisor is a prime number other than 7 (≤105) (\le 10^5), the product of the divisor is less than 1018 10^{18}. The same redundancy has a maximum of 15. Analysis A multiple of 7 in all [L,r] range minus the number that sati

Introduction to China Residue Theorem In Sun Tzu's computing Sutra, there is a problem: "I don't know the number of things today, but the number of three is two (divided by more than three 2 ), three of the five remaining values (divided by more than 5 3), two of the seven remaining values (divided by more than 7 2), and asked ry?" This problem is called the "Sun Tzu's Problem". The general solution of this problem is internationally referred to as t

Chinese remainder theorem againTime limit:1000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): Accepted submission (s): 777Problem description I know some of the students are looking at the Chinese remainder theorem recently, this theorem