HDU——5667Sequence(矩阵快速幂+费马小定理应用)
Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1424 Accepted Submission(s): 469
Problem Description
Holion
August will eat every thing he has found.
Now there are many foods,but he does not want to eat all of them at once,so he find a sequence.
fn=⎧⎩⎨⎪⎪1,ab,abfcn−1fn−2,n=1n=2otherwise
He gives you 5 numbers n,a,b,c,p,and he will eat fn foods.But there are only p foods,so you should tell him fn mod p.
Now there are many foods,but he does not want to eat all of them at once,so he find a sequence.
fn=⎧⎩⎨⎪⎪1,ab,abfcn−1fn−2,n=1n=2otherwise
He gives you 5 numbers n,a,b,c,p,and he will eat fn foods.But there are only p foods,so you should tell him fn mod p.
Input
The
first line has a number,T,means testcase.
Each testcase has 5 numbers,including n,a,b,c,p in a line.
1≤T≤10,1≤n≤1018,1≤a,b,c≤109,p is a prime number,and p≤109+7.
Each testcase has 5 numbers,including n,a,b,c,p in a line.
1≤T≤10,1≤n≤1018,1≤a,b,c≤109,p is a prime number,and p≤109+7.
Output
Output
one number for each case,which is fn mod
p.
Sample Input
1
5 3 3 3 233
Sample Output
190
网上题解解释起来不是非常清楚,我想了一节课终于明白了为什么要对p-1取模了。
首先理解费马小定理:若a与p互质且b为素数,则a^(p-1)%p恒为1。但是这跟题目有啥关系?所以要先推题目的递推式
先两边取loga对数(刚开始取log10,发现化不出来)然后就可以得到。
因此另。即。然后求Kn。(右上角的1与右下边的b互换不影响结果)
然后此题数据有点水,若a为p的倍数(a=x*p),此时原式可以为((x*p)^Kn)%p=0,因此要特判为0,不然结果会是1,由于后台数据并没有考虑这个情况,因此部分人没考虑这个情况直接模p-1也是可以过的,就是这个地方让我纠结了很久:虽然p为素数与绝大部分数都互质,但万一a是p的倍数怎么办?至此得出答案。
代码:
#include<iostream> #include<algorithm> #include<cstdlib> #include<sstream> #include<cstring> #include<cstdio> #include<string> #include<deque> #include<stack> #include<cmath> #include<queue> #include<set> #include<map> #define INF 0x3f3f3f3f #define MM(x) memset(x,0,sizeof(x)) using namespace std; typedef long long LL; LL n,a,b,c,p; struct mat { LL pos[3][3]; mat (){MM(pos);} }; inline mat operator*(const mat &a,const mat &b) { mat c; for (int i=0; i<3; i++) { for (int j=0; j<3; j++) { for (int k=0; k<3; k++) c.pos[i][j]+=(a.pos[i][k]*b.pos[k][j])%(p-1); } } return c; } inline mat operator^(mat a,LL b) { mat r,bas=a; for (int i=0; i<3; i++) r.pos[i][i]=1; while (b!=0) { if(b&1) r=r*bas; bas=bas*bas; b>>=1; } return r; } inline LL qpow(LL a,LL b) { LL r=1,bas=a; while (b!=0) { if(b&1) r=(r*bas)%p; bas=(bas*bas)%p; b>>=1; } return r%p; } int main(void) { int tcase,i,j; scanf("%d",&tcase); while (tcase--) { scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&p); if(a%p==0) puts("0"); else if(n==1) puts("1"); else if(n==2) printf("%lld\n",qpow(a,b)); else { mat t,one; t.pos[0][0]=c;t.pos[0][1]=1;t.pos[0][2]=b; t.pos[1][0]=1; t.pos[2][2]=1; one.pos[0][0]=b;one.pos[1][0]=0;one.pos[2][0]=1; t=t^(n-2); one=t*one; printf("%lld\n",qpow(a,one.pos[0][0])); } } return 0; }