Ipad,IPhone
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 350 Accepted Submission(s): 128
Problem Description
In ACM_DIY, there is one master called “Lost”. As we know he is a “-2Dai”, which means he has a lot of money.
Well, Lost use Ipad and IPhone to reward the ones who solve the following problem.
In this problem, we define F( n ) as :
Then Lost denote a function G(a,b,n,p) as
Here a, b, n, p are all positive integer!
If you could tell Lost the value of G(a,b,n,p) , then you will get one Ipad and one IPhone!
Input
The first line is one integer T indicates the number of the test cases. (T <= 100)
Then for every case, only one line containing 4 positive integers a, b, n and p.
(1 ≤a, b, n, p≤2*10 9 , p is an odd prime number and a,b < p.)
Output
Output one line,the value of the G(a,b,n,p) .
Sample Input
4 2 3 1 10007 2 3 2 10007 2 3 3 10007 2 3 4 10007
Sample Output
40 392 3880 9941
//题意:
给定a,b,n,p,求G(a,b,n,p)=(a^((p-1)/2)+1)*(b^((p-1)/2)+1)*[(sqrt(a)+sqrt(b))^(2*f[n])+(sqrt(a)-sqrt(b)^(2*f[n])]%p的值,其中a,b,n,p,都是正整数;p是奇素数;f[n]是斐波那契数列。
思路:
通过二次剩余的知识可知,x^((p-1)/2)=+-1%p,所以当a,b中至少有一个是模p的二次剩余时,那么表达式的值就是0,否则a,b都是模p的二次剩余,那么就相当于求表达式G(a,b,n,p)=4[(sqrt(a)+sqrt(b))^(2f[n])+(sqrt(a)-sqrt(b))^(2f[n])]%p的值。进一步表示为G(a,b,n,p)=4[(a+b+2sqrt(ab))^f[n]+(a+b-sqrtr(ab))^f[n]]%p,这样就可以得出特征根为a+b+sqrt(ab)和a+b-sqrt(ab),
构造递推矩阵即可求出。
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define ll long long
#define N 2
using namespace std;
struct mat
{
ll m[N][N];
};
mat I=
{
1,0,
0,1
};
mat multi(mat a,mat b,ll p)
{
mat c;
int i,j,k;
for(i=0;i<N;i++)
{
for(j=0;j<N;j++)
{
c.m[i][j]=0;
for(k=0;k<N;k++)
{
c.m[i][j]+=a.m[i][k]*b.m[k][j]%p;
}
c.m[i][j]%=p;
}
}
return c;
}
mat power(mat a,ll k,ll m)
{
mat ans=I,p=a;
while(k)
{
if(k&1)
{
ans=multi(ans,p,m);
k--;
}
k>>=1;
p=multi(p,p,m);
}
return ans;
}
ll quick_mod(ll a,ll b,ll m)
{
ll ans=1;
a%=m;
while(b)
{
if(b&1)
{
ans=ans*a%m;
b--;
}
b>>=1;
a=a*a%m;
}
return ans;
}
int main()
{
int t;
mat res;
ll a,b,n,p;
ll x,y,z,ret;
scanf("%d",&t);
while(t--)
{
mat A={0,1,1,1};
scanf("%lld%lld%lld%lld",&a,&b,&n,&p);
x=quick_mod(a,(p-1)>>1,p);
y=quick_mod(b,(p-1)>>1,p);
if(x!=1||y!=1)
{
printf("0\n");
continue;
}
mat ans=power(A,n,p-1);
z=ans.m[0][0]+ans.m[0][1];
z%=p-1;
res.m[0][0]=2*(a+b)%p;
res.m[0][1]=-(a-b)*(a-b)%p;
res.m[1][0]=1;
res.m[1][1]=0;
ans=power(res,z-1,p);
ret=(ans.m[0][0]*((2*a+2*b)%p)%p+ans.m[0][1]*2%p)%p;
ret=(ret+p)%p;
ret=(ret*4)%p;
printf("%lld\n",ret);
}
return 0;
}