HDU3221Brute-force Algorithm(矩陣快速冪&&指數降冪)

bigbigship發表於2014-07-01
Go矩陣

題目連結：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



寫出前幾項為 a,b,ab,ab^2,a^2b^3,a^3,b^5;



從第三項開始 a的指數為斐波那契數列


由於比較大 我們根據公式a^b%p=a^(b%phi(p)+phi(p))%p   b>=phi(p)對指數進行降冪


然後再快速冪取模即可得到結果


對p=1的情況進行特判；


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



 




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