hdu 3221 Brute-force Algorithm

Brute-force Algorithm

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1825    Accepted Submission(s): 444

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

 

分析：见hdu4550

#include<cstdio>
#include<cmath>
#define ll __int64
ll a,b,p,n;
struct Matrix{
	ll v[2][2];
};
Matrix A={0,1,1,1},B={1,0,0,1};
ll euler(ll x){
	ll i,ans=x,t=(ll)sqrt(x*10.0);
	for(i=2;i<=t;i++)if(x%i==0){
		ans=ans/i*(i-1);
		while(x%i==0)x/=i;
	}
	if(x>1)ans=ans/x*(x-1);
	return ans;
}
ll quickpow(ll x,ll n){
	ll ans=1;
	while(n){
		if(n&1)ans=ans*x%p;
		n>>=1;
		x=x*x%p;
	}
	return ans;
}
Matrix mul(Matrix m1,Matrix m2,ll M){
	int i,j,k;
	Matrix c;
	for(i=0;i<2;i++)for(j=0;j<2;j++){
		c.v[i][j]=0;
		for(k=0;k<2;k++)c.v[i][j]+=(m1.v[i][k]*m2.v[k][j])%M;
		c.v[i][j]%=M;
	}
	return c;
}
Matrix Mpow(int n,ll M){
	Matrix x=A,c=B;
	while(n>=1){
		if(n&1)c=mul(c,x,M);
		n>>=1;
		x=mul(x,x,M);
	}
	return c;
}
int main(){
	int T,ca=1,i;
	scanf("%d",&T);
	while(T--){
		scanf("%I64d%I64d%I64d%I64d",&a,&b,&p,&n);
		printf("Case #%d: ",ca++);
		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%p)*(b%p))%p);continue;}
		ll phi=euler(p),f1=0,f2=1,f3,aa,bb;bool j=0;
		for(i=4;i<=n;i++){
			f3=f1+f2;
			if(f3>=phi){f3%=phi;j=1;break;}
			f1=f2,f2=f3;
		}
		if(j){
			Matrix res=Mpow(n-3,phi);
			f1=res.v[1][0],f2=res.v[1][1],f3=f1+f2;
			aa=f2%phi+phi,bb=f3%phi+phi;
		}
		else {aa=f3,bb=f1+f2;}
		printf("%I64d\n",(quickpow(a,aa)*quickpow(b,bb))%p);
	}
	return 0;
}