标签：scan turn 快速幂 scanf 质数 style ace namespace pen
题目不难懂。式子是一个递推式，并且不难发现f[n]都是a的整数次幂。(f[1]=a0;f[2]=ab;f[3]=ab*f[2]c*f[1]...)
我们先只看指数部分，设h[n]. 则
h[1]=0;
h[2]=b;
h[3]=b+h[2]*c+h[1];
h[n]=b+h[n-1]*c+h[n-1].
h[n]式三个数之和的递推式，所以就可以转化为3x3的矩阵与3x1的矩阵相乘。于是
h[n] c 1 b h[n-1]
h[n-1] = 1 0 0 * h[n-2]
1 0 0 1 1
又根据费马小定理(ap-1%p=1,p是质数且a,p互质)可得：ah[n]%mod=ah[n]%(mod-1)%mod.
因为 ah[n]%mod= ax*(mod-1)+h[n]%(mod-1)%mod = ax*(mod-1)*ah[n]%(mod-1)%mod = ah[n]%(mod-1)%mod;
#include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef long long ll; ll p; struct Mat { ll mat[3][3]; }; Mat Multiply(Mat a, Mat b) { Mat c; memset(c.mat, 0, sizeof(c.mat)); for(int k = 0; k < 3; ++k) for(int i = 0; i < 3; ++i) if(a.mat[i][k]) for(int j = 0; j < 3; ++j) if(b.mat[k][j]) c.mat[i][j] = (c.mat[i][j] +a.mat[i][k] * b.mat[k][j])%(p-1); return c; } Mat QuickPower(Mat a, ll k) { Mat c; memset(c.mat,0,sizeof(c.mat)); for(int i = 0; i <3; ++i) c.mat[i][i]=1; for(; k; k >>= 1) { if(k&1) c = Multiply(c,a); a = Multiply(a,a); } return c; } ll Powermod(ll a,ll b) { a%=p; ll ans=1; for(; b; b>>=1) { if(b&1) ans=(ans*a)%p; a=(a*a)%p; } return ans; } int main() { //freopen("in.txt","r",stdin); int T; scanf("%d",&T); ll n,a,b,c; Mat x; while(T--) { scanf("%I64d%I64d%I64d%I64d%I64d",&n,&a,&b,&c,&p); if(n==1) printf("1\n"); else if(n==2) printf("%I64d\n",Powermod(a,b)); else { x.mat[0][0]=c; x.mat[0][1]=1; x.mat[0][2]=b; x.mat[1][0]=1; x.mat[1][1]=0; x.mat[1][2]=0; x.mat[2][0]=0; x.mat[2][1]=0; x.mat[2][2]=1; x=QuickPower(x,n-2); ll k=(x.mat[0][0]*b+x.mat[0][2]); printf("%I64d\n",Powermod(a,k)); } } return 0; }
标签：scan turn 快速幂 scanf 质数 style ace namespace pen
原文地址：http://www.cnblogs.com/pach/p/5990548.html
