HDU3221Brute-force Algorithm(矩阵快速幂&&指数降幂)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3221

Problem Description

Professor Brute is not good at algorithm design. Once he was asked to solve a path finding problem. He worked on it for several days and finally came up with the following algorithm:



Any fool but Brute knows that the function “funny” will be called too many times. Brute wants to investigate the number of times the function will be called, but he is too lazy to do it.

Now your task is to calculate how many times the function “funny” will be called, for the given a, b and n. Because the answer may be too large, you should output the answer module by P.

 

Input

There are multiple test cases. The first line of the input contains an integer T, meaning the number of the test cases.

For each test cases, there are four integers a, b, P and n in a single line.

You can assume that 1≤n≤1000000000, 1≤P≤1000000, 0≤a, b<1000000.

 

Output

For each test case, output the answer with case number in a single line.

 

Sample Input

3
3 4 10 3
4 5 13 5
3 2 19 100

 

Sample Output

Case #1: 2
Case #2: 11
Case #3: 12

写出前几项为 a,b,ab,ab^2,a^2b^3,a^3,b^5;

从第三项开始 a的指数为斐波那契数列

由于比较大 我们根据公式a^b%p=a^(b%phi(p)+phi(p))%p   b>=phi(p)对指数进行降幂

然后再快速幂取模即可得到结果

对p=1的情况进行特判；

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
const int N = 2;
struct matrax
{
LL m

;
};
matrax E={
1,0,
0,1
};
matrax A={
1,1,
1,0
};
LL a,b,p,n,mm;
matrax multi(matrax a,matrax b)
{
matrax c;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
c.m[i][j]=0;
for(int k=0;k<N;k++)
c.m[i][j]+=a.m[i][k]*b.m[k][j];
if(c.m[i][j]>mm)
c.m[i][j]=c.m[i][j]%mm+mm;
}
}
return c;
}
matrax pow(matrax a,LL n)
{
matrax ans=E,p=a;
while(n){
if(n&1){
ans=multi(ans,p);
n--;
}
n>>=1;
p=multi(p,p);
}
return ans;
}
LL phi(LL n)
{
LL rea=n;
for(int i=2;i*i<=n;i++){
if(n%i==0){
rea=rea-rea/i;
while(n%i==0)
n/=i;
}
}
if(n>1)
rea=rea-rea/n;
return rea;
}
LL quick_mod(LL a,LL b)
{
LL ans=1;
while(b){
if(b&1){
ans=ans*a%p;
b--;
}
b>>=1;
a=a*a%p;
}
return ans;
}
int main()
{
int t=1,cas;
cin>>cas;
while(cas--){
cin>>a>>b>>p>>n;
mm=phi(p);
printf("Case #%d: ",t++);
if(n==1){
printf("%I64d\n",a%p);
continue;
}
if(n==2){
printf("%I64d\n",b%p);
continue;
}
if(n==3){
printf("%I64d\n",a*b%p);
continue;
}
if(p==1){
puts("0");
continue;
}
matrax g=pow(A,n-2);
LL m1,m2,num1,num2;
m1=g.m[1][0];
m2=g.m[0][1]+g.m[1][1];
if(m2>mm) m2=m2%mm+mm;
num1=quick_mod(a,m1);
num2=quick_mod(b,m2);
printf("%I64d\n",num1*num2%p);
}
return 0;
}
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
const int N = 2;
struct matrax
{
LL m

;
};
matrax E={
1,0,
0,1
};
matrax A={
1,1,
1,0
};
LL a,b,p,n,mm;
matrax multi(matrax a,matrax b)
{
matrax c;
for(int i=0;i<N;i++){
for(int j=0;j<N;j++){
c.m[i][j]=0;
for(int k=0;k<N;k++)
c.m[i][j]+=a.m[i][k]*b.m[k][j];
if(c.m[i][j]>mm)
c.m[i][j]=c.m[i][j]%mm+mm;
}
}
return c;
}
matrax pow(matrax a,LL n)
{
matrax ans=E,p=a;
while(n){
if(n&1){
ans=multi(ans,p);
n--;
}
n>>=1;
p=multi(p,p);
}
return ans;
}
LL phi(LL n)
{
LL rea=n;
for(int i=2;i*i<=n;i++){
if(n%i==0){
rea=rea-rea/i;
while(n%i==0)
n/=i;
}
}
if(n>1)
rea=rea-rea/n;
return rea;
}
LL quick_mod(LL a,LL b)
{
LL ans=1;
while(b){
if(b&1){
ans=ans*a%p;
b--;
}
b>>=1;
a=a*a%p;
}
return ans;
}
int main()
{
int t=1,cas;
cin>>cas;
while(cas--){
cin>>a>>b>>p>>n;
mm=phi(p);
printf("Case #%d: ",t++);
if(n==1){
printf("%I64d\n",a%p);
continue;
}
if(n==2){
printf("%I64d\n",b%p);
continue;
}
if(n==3){
printf("%I64d\n",a*b%p);
continue;
}
if(p==1){
puts("0");
continue;
}
matrax g=pow(A,n-2);
LL m1,m2,num1,num2;
m1=g.m[1][0];
m2=g.m[0][1]+g.m[1][1];
if(m2>mm) m2=m2%mm+mm;
num1=quick_mod(a,m1);
num2=quick_mod(b,m2);
printf("%I64d\n",num1*num2%p);
}
return 0;
}