2015
04-14

# Evil teacher’s Final Problem

In the math class, the evil teacher gave you one unprecedented problem!

Here f(n) is the n-th fibonacci number (n >= 0)! Where f(0) = f(1) = 1 and for any n > 1, f(n) = f(n – 1) + f(n – 2). For example, f(2) = 2, f(3) = 3, f(4) = 5 …

The teacher used to let you calculate f(n) mod p where n <= 10^18 and p <= 10^9, however , as an ACMER, you may just kill it in seconds! The evil teacher is mad about this. As you kill the Evil teacher.s problem in second too!!! now he let you calculate G(n,k) .Here G(n,0) = f(n) , G(n,i) = f( G(n,i-1) ) (k >= i >= 1). However the G(n,k) may be so large ,so you just need to output the remainder of the answer after divided by p.

Note: This problem is the evil teacher’s final problem, it is really hard ! If you could solve this problem during the competition, you will be reward in the ACM_DIY gathering.

The first line is one integer T indicates the number of the test cases. (T <=500000)

Then for every case, three integers n k and p . (0 <= n <= 10^9,0 <= k <= 10^4, 1 <= p <= 10^8)

The first line is one integer T indicates the number of the test cases. (T <=500000)

Then for every case, three integers n k and p . (0 <= n <= 10^9,0 <= k <= 10^4, 1 <= p <= 10^8)

10
1 10000 100000000
2 3 10000
3 97 98
4 2 10
5 1 29
1234 5678 100000000
3344 5566 77889900
10000 10000 100000000
1111 10000 90000000
999 876 10000000

Case #1: 1
Case #2: 2
Case #3: 3
Case #4: 4
Case #5: 5
Case #6: 17835789
Case #7: 5381861
Case #8: 71647609
Case #9: 43710337
Case #10: 9102595

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define maxn 20050
bool isprime[maxn];
int prime[maxn],nprime,mod[maxn],M;
int primeloop[maxn];
void getprime()
{
long long i,j;
memset(isprime,1,sizeof(isprime));
nprime=0;
for(i=2; i<maxn; i++)
if(isprime[i])
{
prime[nprime++]=i;
for(j=i*i; j<maxn; j+=i) isprime[j]=0;
}
}
int factor[100][2],tol,fac[100][2],numfac;
void findfac(int n,int f[][2],int &t)
{
int x=n,l=(int)sqrt(1.0*n);
t=0;
for(int i=0; i<50; i++) f[i][0]=f[i][1]=0;
for(int i=0; prime[i]<=l; i++)
if(x%prime[i]==0)
{
f[t][0]=prime[i];
while(x%prime[i]==0) f[t][1]++,x/=prime[i];
t++;
}
if(x>1) f[t][0]=x,f[t++][1]++;
}
const int MAX=2;
typedef struct
{
long long m[MAX][MAX];
} Matrix;
Matrix P,I;
Matrix matrixmul(Matrix a,Matrix b) //矩阵乘法
{
int i,j,k;
Matrix c;
for (i=0; i<MAX; i++)
for (j=0; j<MAX; j++)
{
c.m[i][j]=0;
for(k=0; k<MAX; k++)
c.m[i][j]+=((a.m[i][k]%M)*(b.m[k][j]%M))%M;
c.m[i][j]%=M;
}
return c;
}
void quickpow(int n,int &x,int &y)
{
Matrix m=P,b=I;
while(n)
{
if(n&1) b=matrixmul(b,m);
n>>=1,m=matrixmul(m,m);
}
x=b.m[0][0],y=b.m[1][0];
}
int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
}
int exp_mod(int a,int b,int c)
{
int ans=1;
a%=c;
while(b)
{
if(b&1) ans=ans*a%c;
b>>=1,a=a*a%c;
}
return ans;
}
int minloop,ff[maxn],numff;
void dfs(int num,int s=1)
{
if(num==numfac)
{
ff[numff++]=s;
return;
}
for(int i=0; i<=fac[num][1]; i++)
dfs(num+1,s),s*=fac[num][0];
}
int getPrimeLoop(int p)
{
if(p==2) return 3;
if(p==3) return 8;
if(p==5) return 20;
M=p;
if(exp_mod(5,(p-1)>>1,p)==1) p--;
else p=2*p+2;
findfac(p,fac,numfac);
minloop=1e9;
numff=0;
dfs(0,1);
sort(ff,ff+numff);
int x,y;
for(int i=1; i<numff; i++)
{
quickpow(ff[i]-1,x,y);
if(x==0&&y==1) return ff[i];
}
}
int getLoop(int p,int k)
{
int ret;
if(p>19583)
ret=getPrimeLoop(p);
else
ret=primeloop[p];
for(int i=0; i<k-1; i++) ret*=p;
return ret;
}
int getmod(int n)
{
findfac(n,factor,tol);
int ret=1,tem;
for(int i=0; i<tol; i++)
tem=getLoop(factor[i][0],factor[i][1]),ret=ret/gcd(ret,tem)*tem;
return ret;
}
int main()
{
int t,ca=0,p,n,k,x,y;
getprime();
P.m[0][0]=P.m[0][1]=P.m[1][0]=1,P.m[1][1]=0;
I.m[0][0]=I.m[1][1]=1,I.m[1][0]=I.m[0][1]=0;
for(int i=0; i<2250; i++)
primeloop[prime[i]]=getPrimeLoop(prime[i]);
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&k,&p);
mod[0]=p;
for(int i=1; i<=k; i++) mod[i]=getmod(mod[i-1]);
for(int i=k; i>=0; i--)
{
if(i<k) n%=mod[i+1];
M=mod[i],quickpow(n,x,y),n=x;
}
printf("Case #%d: %d\n",++ca,n);
}
return 0;
}


1. 这道题目的核心一句话是：取还是不取。
如果当前取，则index+1作为参数。如果当前不取，则任用index作为参数。

2. 换句话说，A[k/2-1]不可能大于两数组合并之后的第k小值，所以我们可以将其抛弃。
应该是，不可能小于合并后的第K小值吧