HDU 2855解题报告

Fibonacci Check-up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1198    Accepted Submission(s): 675

Problem Description

Every ALPC has his own alpc-number just like alpc12, alpc55, alpc62 etc.
As more and more fresh man join us. How to number them? And how to avoid their alpc-number conflicted? 
Of course, we can number them one by one, but that’s too bored! So ALPCs use another method called Fibonacci Check-up in spite of collision. 

First you should multiply all digit of your studying number to get a number n (maybe huge).
Then use Fibonacci Check-up!
Fibonacci sequence is well-known to everyone. People define Fibonacci sequence as follows: F(0) = 0, F(1) = 1. F(n) = F(n-1) + F(n-2), n>=2. It’s easy for us to calculate F(n) mod m. 
But in this method we make the problem has more challenge. We calculate the formula , is the combination number. The answer mod m (the total number of alpc team members) is just your alpc-number.

 

Input

First line is the testcase T.
Following T lines, each line is two integers n, m ( 0<= n <= 10^9, 1 <= m <= 30000 )

 

Output

Output the alpc-number.

 

Sample Input

21 300002 30000

 

Sample Output

13

 

Source

2009 Multi-University Training Contest 5 - Host by NUDT

 

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        参考代码：

     

#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<string>#include<vector>#include<map>#include<set>#include<stack>#include<queue>#include<ctime>#include<cstdlib>#include<iomanip>#include<utility>#define pb push_back#define mp make_pair#define CLR(x) memset(x,0,sizeof(x))#define _CLR(x) memset(x,-1,sizeof(x))#define REP(i,n) for(int i=0;i<n;i++)#define Debug(x) cout<<#x<<"="<<x<<" "<<endl#define REP(i,l,r) for(int i=l;i<=r;i++)#define rep(i,l,r) for(int i=l;i<r;i++)#define RREP(i,l,r) for(int i=l;i>=r;i--)#define rrep(i,l,r) for(int i=1;i>r;i--)#define read(x) scanf("%d",&x)#define put(x) printf("%d\n",x)#define ll long long#define lson l,m,rt<<1#define rson m+1,r,rt<<11using namespace std;int t,n,m;struct mat{    int d[2][2];}A,B,E;mat multi(mat a,mat b){    mat ans;    rep(i,0,2)    {        rep(j,0,2)        {            ans.d[i][j]=0;            rep(k,0,2)                ans.d[i][j]+=a.d[i][k]*b.d[k][j];            ans.d[i][j]%=m;        }    }    return ans;}mat quickmulti(mat a,int n){    if(n==0) return E;    if(n==1) return a;    mat ans=E;    while(n)    {        if(n&1)        {            n--;            ans=multi(ans,a);        }        else        {            n>>=1;            a=multi(a,a);        }    }    return ans;}int main(){   read(t);   E.d[0][0]=E.d[1][1]=1;E.d[0][1]=E.d[1][0]=0;   while(t--)   {       read(n);read(m);       A.d[0][0]=2;A.d[0][1]=A.d[1][0]=A.d[1][1]=1;       B.d[0][0]=1,B.d[0][1]=B.d[1][0]=B.d[1][1]=0;       mat ans=quickmulti(A,n);       ans=multi(ans,B);       printf("%d\n",ans.d[1][0]);   }}