fast math com

A simple math problem Time Limit: 3000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)Total submission (s): 2780 accepted submission (s): 1649 Problem descriptionlele now is thinking about a simple function f (x ). If x If X> = 10 f (x) = A0 * F (x-1) + A1 * F (X-2) + A2 * F (X-3) + ...... + A9 * F (X-10 ); And AI (0 Now, I will give a0 ~ A9 and two positive integers K and M, and cocould you help Lele to caculate F (k) % m. Inputth

Hdu 1757 A Simple Math Problem (matrix fast power)A Simple Math ProblemTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission (s): 2512 Accepted Submission (s): 1461Problem DescriptionLele now is thinking about a simple function f (x ).If x If x> = 10 f (x) = a0 * f (x-1) + a1 * f (X-2) + a2 * f (X-3) + ...... + A9 * f (X

2326: [HNOI2011] Math job time limit:10 Sec Memory limit:128 MBsubmit:1388 solved:799[Submit] [Status] [Discuss] DescriptionInput Output sample input sample output HINT SourceMatrix multiplication#include #include#include#include#include#include#definell Long Longusing namespacestd;ll n,m,a[4][4],b[4][4];ll mu (ll x,ll y) {ll s=0; while(y) {if(y1) s= (s+x)%m; X= (x1)%m; Y>>=1; } returns;}voidMul (LL a[4][4],ll b[4][4],ll ans[4][4]) {ll t[

mod (b)//expand to find the inverse element//O (LOGN)voidEXGCD (ll A, ll B, LL x, LL y, LL d) { if(b = =0) {x=1; Y=0; D=A; } Else{EXGCD (b, a%b, y, X, D); Y-= a/b*x; }}//a = BX (mod N)ll Moddiv (ll A, ll b) {ll x, Y, D; EXGCD (b, MOD, X, y, D); X= (x+mod)%MOD; X= (x*a/d)%MOD; returnx;}//Fast Power M^nll Quickpow (ll X, ll N) {ll a=1; while(n) {a*= n1? X:1; A%=MOD; N>>=1 ; X*=x; X%=MOD; } returnA;}voidWork () {LL now=B1, TMP; now%=MOD; f

Topic Link: Poke MeMain topic:A function f (x) when x When x >= 10 o'clock, f (x) = a0 * F (X-1) + A1 * F (x-2) + A2 * F (x-3) + ... + A9 * F (x-10);For K and M, ask for the value f (k)% mSample explanation:SlightlyProblem Solving Ideas:A matrix of 10 * 10 is established with a matrix fast power, as follows0 1 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 00 0 0 1 0 0 0 0 0 00 0 0 0 1 0 0 0 0 00 0 0 0 0 1 0 0 0 00 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 1 0 00 0 0 0 0 0 0

three bit better think, direct fast power to take surplus on it. The primary three-bit calculation.We can think of a number, for example 1589. If this number is converted, replace it with 10^x. Then x is equal to LOG10 (1589). And if we put log10 (1589) Mod 1, the result of the 1 remainder is that we only take the decimal digits of x. Imagine, if the x is divided into integers z and fractional g, then 10^x=10^ (z+g) =10^g*10^z,z is an integer, so 10^

Topic:A Simple Math problemTime limit:3000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)Total submission (s): 3522 Accepted Submission (s): 2130Problem Descriptionlele Now was thinking about a simple function f (x).If x If x >= f (x) = a0 * F (X-1) + A1 * F (x-2) + A2 * F (x-3) + ... + A9 * F (x-10);and AI (0Now, I'll give A0 ~ A9 and positive integers k and M, and could you help Lele to Caculate f (k)%m.Inputthe problem contains mut

(a, B) = c --> a * B = gcd (a, B) * C; --> A/gcd (a, B) = C/B, because a/gcd (a, B) must be an integer, so B must be a factor of C, enumerate the factors of C. At first, the pure brute force enumeration of the factor T of C was triggered to understand that mathematics was king. The enumeration factor has been optimized when determining the prime number, that is, it only needs to be enumerated to SQRT (c ). Another condition is that a must be a factor of C. Because b/gcd(a,b)==c/a; #include U

Question Link Question: F (k) % m Thought: f (x) = A0 * F (x-1) + A1 * F (X-2) + A2 * F (X-3) + ...... + A9 * F (X-10), so we can get a Matrix(A0, A1, A2, A3, A4, A5, A6, A7, A8, A9)(1, 0, 0, 0, 0, 0, 0, 0, 0)(0, 1, 0, 0, 0, 0, 0, 0, 0)(0, 0, 1, 0, 0, 0, 0, 0, 0, 0)(0, 0, 0, 1, 0, 0, 0, 0, 0)(0, 0, 0, 0, 1, 0, 0, 0, 0)(0, 0, 0, 0, 0, 1, 0, 0, 0)(0, 0, 0, 0, 0, 0, 1, 0, 0)(0, 0, 0, 0, 0, 0, 0, 1, 0, 0)(0, 0, 0, 0, 0, 0, 0, 0, 1, 0 )*| F (x-1), F (x-2), F (x-3), F (x-4), F (x-5 ), f (x-6), F (x-

incidentally, for the Dirichlet convolution, there is a single-element E.namely: F*e=f,This unitary function has a property e[1]=1,e[k]=0,k=1,2,3 ...This will be used when the power is fast.And then on the code#include #includeusing namespaceStd;typedefLong LongLL;Const intn=1e5+5;ConstLL mod=1e9+7;intn,k; LL F[n],s[n],ans[n],tmp[n];voidsolve () { while(k) {if(k1) { for(intI=1; i0; for(intI=1; i*ii) {tmp[i*i]= (Tmp[i*i]+ans[i]*s[i])%MoD; for(intj=i+1; j*ij) Tmp[i*j]+= (Ans[

1#include 2#include 3#include 4#include 5 #defineRep (i,l,r) for (int i=l;i6 #defineCLR (a,x) memset (A,x,sizeof (a))7 #defineINF 1000038 using namespacestd;9typedefLong Longll;Ten ll N,m; One intMain () A { -Cin>>m>>N; -ll sum=m,ans=m,f=m,k=m-1; the--N; - while(n) - { - if(n1){ +(sum*=f)%=inf; -(ans*=k)%=inf; + } A(k*=k)%=inf; at(f*=f)%=inf; -n>>=1; - } -printf"%lld", (Sum-ans+inf)%inf); - return 0; -}View Code 1008: [HNOI2008] Jailbreak time limit:1

A Simple Math problemTime limit:MS Memory Limit:32768KB 64bit IO Format:%i64d %i64u SubmitStatusPracticeHDU 1757Appoint Description:System Crawler (2015-03-04)DescriptionLele now was thinking about a simple function f (x).If x If x >= f (x) = a0 * F (X-1) + A1 * F (x-2) + A2 * F (x-3) + ... + A9 * F (x-10);and AI (0Now, I'll give A0 ~ A9 and positive integers k and M, and could you help Lele to Caculate f (k)%m.InputThe problem contains mutiple test

(A.BC * 10^m)M is the integer portion of k * LG (N), and LG (A.BC) is a fractional part of K * LG (N)x = LG (A.BC) = k * LG (n)-m = k * LG (N)-(int) (k * LG (N))A.BC = POW (ten, X);ABC = A.BC * 100;So the first three digits of ABC can find#include #include#includestring.h>#include#includeusing namespaceStd;typedefLong Longll;intPow (intAintb) { intAns =1; A%= +; while(b) {if(b%2!=0) ans= (ans * a)% +; A= (A * a)% +; b/=2; } returnans;}//Fast n

[HDU 1757] A Simple Math Problem (matrix fast power) Question: If x If x> = 10 f (x) = a0 * f (x-1) + a1 * f (X-2) + a2 * f (X-3) + ...... + A9 * f (X-10 );And ai (0 For k and m, evaluate f (k) % m. Solution: F (x) = a0 * f (x-1) + a1 * f (X-2) + a2 * f (X-3) + ...... + A9 * f (X-10) K is 10 ^ 9 orders of magnitude, so it cannot be done in recursive mode. This type of question can be done by constructing a

#include using namespace STD;typedef Long LongLL;Const intn=Ten;intKt,m;structmatrix{intMp[n][n];}; Matrix Mul (Matrix A,matrix b) {intI,j,k; Matrix C; for(i=0; iTen; i++) for(j=0; jTen; J + +) {c.mp[i][j]=0; for(k=0; kTen; k++) {c.mp[i][j]= (c.mp[i][j]+a.mp[i][k]*b.mp[k][j])%m; } }returnC;} MatrixPOW(Matrix T,intx) {if(x==1)returnT Matrix C; C=POW(t,x/2); C=mul (C,C);if(x1)returnMul (c,t);Else returnC;}intMain () {intI,j,ans; Matrix a,t; while(~scanf("%d%d", kt,m)) {

Topic Link: Click to open the linkTest instructions: n^m%1000000007 N (1 ,m (1 A little pit. n Too much can overflow, pow_mod (n,m,mod) =pow (n%mod,m,mod)Deduce it ...N^m%mod= (n%mod+k*mod) ^m%mod=[(n%mod) ^m +: A bunch of mods]%mod = (n%mod) ^m%modHaven't hit the code for ages .... QaqHardware to get out of the Xiang ... All day long hold the video to see how people write how to write ... Various circuit diagrams fly in the sky ..... It's like a yard farm.The fact proves that the algorithm is

Address: HDU 1757 Finally, the matrix will be constructed. In fact, it is not difficult, just blame yourself for being stupid .. =! F (x) = A0 * F (x-1) + A1 * F (X-2) + A2 * F (X-3) + ...... + A9 * F (X-10)The constructed matrix is: (The Matrix constructed in my code is reversed up and down)| 0 1 0... 0 | f0 | F1 || 0 0 1 0... 0 | F1 | F2 || ...... 1 | * |... | = || A9 A8 ...... A0 | F9 | F10 | Then, according to the combination Law of the matrix, we can first calculate the (K-9) Power of the c

If x If X> = 10 f (x) = A0 * F (x-1) + A1 * F (X-2) + A2 * F (X-3) + ...... + A9 * F (X-10 ); For such a formula, you can easily construct such a matrix through the relationship between the matrix and linear transformation. A0: 9876543210 A1:1 1 1 1 1 1 1 1 1 11 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 00 0 0 1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 00 0 0 0 0 0 1 0 00 0 0 0 0 0 0 1 0 Then f (n) = A1 ^ (N-9) * a0 Quick power #include 　　 HDU 1757 a simple