hdoj 5895 Mathematician QSC 【数论----矩阵快速幂求解类斐波那契数列】
Mathematician QSC
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 342 Accepted Submission(s): 184
Problem Description
QSC dream of becoming a mathematician, he believes that everything in this world has a mathematical law.
Through unremitting efforts, one day he finally found the QSC sequence, it is a very magical sequence, can be calculated by a series of calculations to predict the results of a course of a semester of a student.
This sequence is such like that, first of all, f(0)=0,f(1)=1,f(n)=f(n−2)+2∗f(n−1)(n≥2)Then the definition of the QSC sequence is g(n)=∑ni=0f(i)2. If we know the birthday of the student is n, the year at the beginning of the semester is y, the course number x and the course total score s, then the forecast mark is xg(n∗y)%(s+1).
QSC sequence published caused a sensation, after a number of students to find out the results of the prediction is very accurate, the shortcoming is the complex calculation. As clever as you are, can you write a program to predict the mark?
Through unremitting efforts, one day he finally found the QSC sequence, it is a very magical sequence, can be calculated by a series of calculations to predict the results of a course of a semester of a student.
This sequence is such like that, first of all, f(0)=0,f(1)=1,f(n)=f(n−2)+2∗f(n−1)(n≥2)Then the definition of the QSC sequence is g(n)=∑ni=0f(i)2. If we know the birthday of the student is n, the year at the beginning of the semester is y, the course number x and the course total score s, then the forecast mark is xg(n∗y)%(s+1).
QSC sequence published caused a sensation, after a number of students to find out the results of the prediction is very accurate, the shortcoming is the complex calculation. As clever as you are, can you write a program to predict the mark?
Input
First line is an integer T(1≤T≤1000).
The next T lines were given n, y, x, s, respectively.
n、x is 8 bits decimal integer, for example, 00001234.
y is 4 bits decimal integer, for example, 1234.
n、x、y are not negetive.
1≤s≤100000000
The next T lines were given n, y, x, s, respectively.
n、x is 8 bits decimal integer, for example, 00001234.
y is 4 bits decimal integer, for example, 1234.
n、x、y are not negetive.
1≤s≤100000000
Output
For each test case the output is only one integer number ans in a line.
Sample Input
2 20160830 2016 12345678 666 20101010 2014 03030303 333
Sample Output
1 317
Source
2016 ACM/ICPC Asia Regional Shenyang Online
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推出公式：
1. F（n）= G（n）*G（n+1）/2:；
2. F ( n）= 5* F(n-1) + 5* F（n-2）-F（n-3）
公式：
(x^n)%c=x^(n%phi(c)+phi(c))%c;
原则：
方阵有结合律---没有交换律
思路：
先求出phi（s+1）--然后用矩阵求出F（n）%phi+phi---
然后快速幂求答案---
构造矩阵
A:
5 5 -1
1 0 0
0 1 0
和列向量
B
F [ 2 ]
F [ 1 ]
F [ 0 ]
C =A*B=
F [ 3 ]
F [ 2 ]
F [ 1 ]
F [ n ] F ( n-1 ) F [ 2 ]
F [ n-1] = A * F ( n-2 ) =A...........=A^(n-2)* F [ 1 ]
F [n-2] F (n-3) F [ 0 ]
代码1：
注意：
F（n）= G（n）*G（n+1）/2:
因为最后要除2---所以取模时我们要对2*phi取模-----
#include
#include
#include
using namespace std;
#define LLL long long
LLL mod;
struct node{
LLL shu[2][2];
void clear()
{
for (int i=0;i<2;i++)
for (int j=0;j<2;j++)
shu[i][j]=0;
}
node operator * (const node &B) const
{
node C;
C.clear();
for (int i=0;i<2;i++)
for (int j=0;j<2;j++)
for (int k=0;k<2;k++)
C.shu[i][j]=(C.shu[i][j]+shu[i][k]*B.shu[k][j])%mod;
return C;
}
}A,B,C,D;
LLL P[2]={0,1};
LLL phi(LLL xx)
{
LLL lp=xx;
for (LLL i=2;i*i<=xx;i++)
{
if (xx%i==0)
{
lp=lp-lp/i;
while (xx%i==0)
xx/=i;
}
}
if (xx>1)
lp=lp-lp/xx;
return lp;
}
LLL g(LLL xx)
{
LLL lp;
mod*=2;//注意
A.clear();
for (int i=0;i<2;i++)
A.shu[i][i]=1;
B.clear();
B.shu[1][0]=B.shu[0][1]=1;
B.shu[0][0]=2;
if (xx<3)
return P[xx];
while (xx)
{
if (xx&1)
A=A*B;
B=B*B;
xx>>=1;
}
lp=0;
lp=(A.shu[0][0]*A.shu[0][1])/2;
mod/=2;//注意
lp+=mod;
return lp;
}
void slove(LLL x,LLL n,LLL s)
{
mod=phi(s);
LLL lp=g(n);
LLL ans=1;
while (lp)
{
if (lp&1)
ans=(ans*x)%s;
x=(x*x)%s;
lp>>=1;
}
printf("%lld\n",ans);
}
int main()
{
int t;scanf("%d",&t);
while (t--)
{
LLL n,y,x,s;
scanf("%lld%lld%lld%lld",&n,&y,&x,&s);
n*=y;
s+=1;
slove(x,n,s);
}
return 0;
}
代码：
#include
#include
#include
using namespace std;
#define LLL long long
LLL mod;
struct node{
LLL shu[3][3];
void clear()
{
for (int i=0;i<3;i++)
for (int j=0;j<3;j++)
shu[i][j]=0;
}
node operator * (const node &B) const
{
node C;
C.clear();
for (int i=0;i<3;i++)
for (int j=0;j<3;j++)
for (int k=0;k<3;k++)
C.shu[i][j]=(C.shu[i][j]+shu[i][k]*B.shu[k][j])%mod;
return C;
}
}A,B,C,D;
LLL P[3]={0,1,5};
LLL phi(LLL xx)
{
LLL lp=xx;
for (LLL i=2;i*i<=xx;i++)
{
if (xx%i==0)
{
lp=lp-lp/i;
while (xx%i==0)
xx/=i;
}
}
if (xx>1)
lp=lp-lp/xx;
return lp;
}
LLL g(LLL xx)
{
LLL lp;
A.clear();
for (int i=0;i<3;i++)
A.shu[i][i]=1;
B.clear();
B.shu[0][0]=B.shu[0][1]=5;B.shu[0][2]=-1;
B.shu[1][0]=B.shu[2][1]=1;
if (xx<3)
return P[xx];
xx-=2;
while (xx)
{
if (xx&1)
A=A*B;
B=B*B;
xx>>=1;
}
lp=0;
for (int i=0;i<3;i++)
lp=(lp+A.shu[0][i]*P[2-i])%mod;
lp+=mod;
return lp;
}
void slove(LLL x,LLL n,LLL s)
{
mod=phi(s);
LLL lp=g(n);
LLL ans=1;
//printf("%lld %lld %lld %lld\n",x,lp,s,mod);
while (lp)
{
if (lp&1)
ans=(ans*x)%s;
x=(x*x)%s;
lp>>=1;
}
printf("%lld\n",ans);
}
int main()
{
int t;scanf("%d",&t);
while (t--)
{
LLL n,y,x,s;
scanf("%lld%lld%lld%lld",&n,&y,&x,&s);
n*=y;
s+=1;
// printf("%lld %lld %lld\n",x,n,s);
slove(x,n,s);
}
return 0;
}