Fast Power + Fermat theorem for CSU1516 matrices

　　Through this problem really learned a lot of things, at least the matrix fast power is a primer, the common fast power also thoroughly understand (formerly are playing template), understand the Fermat theorem

Key (A^fib[b])%p values where p is a prime number, 0<a<p, B is within the INT range

Let's just assume that fib[b]>p-1

(A^fib[b])%p = (a^ (p-1) *a^ (p-1) *....*a^ (p-1) *a^m)%p

(Here m = fib[b]% (p-1) )

Since P is a prime number and a<p so gcd (a,p) = 1, so the fee is very low (a^ (p-1))%p=1

The above formula = (a^ (fib[b]% (p-1)))%p

Then fib[b]% (p-1) can be obtained by the Matrix fast Power

At last, we can get (a^m)%p with the common fast power.

The rest is just an interval DP.

In a hurry to write, the code is very ugly, ll and int everywhere definition ...

Here is the result of the operation, time limit of 2s, a little risk. There are also 1s of ...

3

105459

Round_0

1572 KB

1860 MS

C++

2109 B

2015-04-06 01:51:32

1#include <cstdio>2#include <cstring>3#include <iostream>4#include <algorithm>5 using namespacestd;6 Const intMAXN = the;7 Const intINF =0x3f3f3f3f;8typedefLong LongLL;9 intA[MAXN],B[MAXN];Ten intN; One LL p; A LL SUM[MAXN],DP[MAXN][MAXN]; - structnode - { theLL a[2][2]; - node () -     { -a[0][0] = a[1][0] = a[0][1] =1; +a[1][1] =0; -     } + }; A node Mul (node A,node B) at { - node ans; -      for(inti =0;i<2; ++i) for(intj =0;j<2;++j) -     { -ANS.A[I][J] =0; -          for(intK =0;k<2; ++k) Ans.a[i][j] = (ans.a[i][j]+a.a[i][k]*b.a[k][j])% (P-1); in     } -     returnans; to } + ll Mypow (ll x,ll m,ll MoD) - { theLL ret =1; *      while(m) $     {Panax Notoginseng         if(m&1) ret = (ret*x)%MoD; -x = (x*x)%MoD; theM/=2; +     } A     returnret; the } + intFibintx) - { $ node ans,s; $      while(x) -     { -         if(x&1) ans =Mul (ans,s); thes =Mul (s,s); -X/=2;Wuyi     } the     returnans.a[1][0]% (P-1); - } Wu intSolveintx) - { About     intm = fib (b[x]-1); $     returnMypow (a[x],m,p); - } - ll gcd (ll X,ll y) - { A     returnY?GCD (y,x%y): x; + } the intMain () - { $ //freopen ("In.txt", "R", stdin); the     intKase =1; the      while(cin>>n>>p) the     { the          for(inti =1; i<=n;++i) for(intj =1; j<=n;++j) Dp[i][j] = i==j?0: INF; -          for(inti =1; i<=n;++i) scanf ("%d", A +i); in          for(inti =1; i<=n;++i) scanf ("%d", B +i); the          for(inti =1; i<=n;++i) A[i] =solve (i); the  Aboutsum[0] =0; sum[1] = a[1]; the          for(inti =2; i<=n;++i) Sum[i] = sum[i-1]+(LL) a[i]; theprintf"Case %d:", kase++); the         if(n==1) {printf ("0\n");Continue;} +          for(intR =2; r<=n;++r) for(inti =1; i<=n-r+1;++i) -         { the             intj = i+r-1;Bayi              for(intK = i;k<j;++k) theDp[i][j] = min (dp[i][j],dp[i][k]+dp[k+1][J]+GCD (sum[k]-sum[i-1],sum[j]-sum[k-1])); the         } -cout<<dp[1][n]<<Endl; -     } the     return 0; the}

　　

Fast Power + Fermat theorem for CSU1516 matrices