mod gta

Title Link: https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1185Test instructions: Chinese question eh ~Idea: Wythoff template problem, and 51nod1072 basically the same (http://www.cnblogs.com/geloutingyu/p/6198094.html), but the data is relatively large (1e18), there will be precision problems;We can:Order: Cnt=abs (x-y);Geloutingyu=1e9;a[3]={618033988, 749894848, 204586834} ((sqrt (5) +1)/2=1.618033988749894848204586834, we can first not calculate 1, and finally add a CNT on the

, when the side length is (a, B), a total of (N'-a) * (N'-B) * (gcd (a, B)-1) is invalid. However, because the range of N is too large, the length of the enumeration side is O (n ^ 2) Time-out. Therefore, we need to optimize the solution process: The specific method is to enumerate the GCD value. Because the GCD value is at least 2, it will affect the result. Therefore, the GCD value can be enumerated from 2 and the maximum GCD value is n. If the enumerated GCD value is K, then if there is gcd

After the last summary of the pigeon has not long in fact is about to start school hurriedly to the liver two articlesToday's content--congruence equation and extended Euclidean algorithmCo-yuCongruence definition: If there are two integers, a, a, so that (a-B) mod p is 0, then the "a" is the case with the mod pIn other words, a mod p is equal to the B

"The main topic"If you use F[i][j] to represent the elements of column I, J, in the Matrix, then F[i][j] satisfies the following recursion:F[1][1]=1F[i,j]=a*f[i][j-1]+b (j!=1) ①F[i,1]=c*f[i-1][m]+d (i!=1) ②The a,b,c,d in a recursive style is a given constant. F[N][M].IdeasGrind for a morning, however the extra data on the UOJ is not over yet. Bzoj on the AC first put up, follow slowly grind ...* There is a point, the final output of the answer when the first +

Topic linksThe value of C (n, m)%p, N, mFirst, the value of the C (n, m)%PI is calculated, and then this is an equation of congruence. Solved by the Chinese remainder theorem.#include #include#include#include#includeusing namespacestd;#definell Long Longll a[ -], b[ -];voidExtend_euclid (ll A, ll B, ll x, LL y) { if(b = =0) {x=1; Y=0; return; } extend_euclid (b, a%b, x, y); LL TMP=x; X=y; Y= tmp-(A/b) *y;} LL Mul (ll A, ll N, ll MoD) {a= (a%

, and the modulus (the amount of "overflow") is 12. Now we move the hour hand counterclockwise from 4 to 2, and move clockwise 10 to get the same result. The arithmetic formula is 4-2 and 4+10, at a glance 4-2 is obviously not equal to 4+10, then why the two results on the clock is the same?That's "modulo" in mischief. When the result of the operation exceeds the counting range, the modulo operation is performed, and the 4+10=14>12,12 is 2=4+ (-2) after the modulo is obtained. At this point you

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