HDU 5667 Sequence【矩陣快速冪+費馬小定理】
阿新 • • 發佈:2019-02-03
題目連結:
題意:
Lcomyn 是個很厲害的選手,除了喜歡寫17kb+的程式碼題,偶爾還會寫數學題.他找到了一個數列:
給定各個數,求
分析:
可以發現最後都是a的倍數,這樣我們讓
注意:
- 由費馬小定理可知,
ab%p=ab/(p−1)∗(p−1)+b%(p−1)%p=ab/(p−1)∗(p−1)%p∗ab%(p−1)%p=ab%(p−1)%p ,所以矩陣快速冪的模應該為p 。 - 特別注意
a%p==0 的時候,答案應該為0。
程式碼:
#include<cstdio>
const int N = 105;
int mod = 1e9 + 7;
struct Matrix
{
int row,cal;
long long m[N][N];
};
Matrix init(Matrix a, long long t)
{
for(int i = 0; i < a.row; i++)
for(int j = 0; j < a.cal; j++)
a.m[i][j] = t;
return a;
}
Matrix mul(Matrix a,Matrix b)
{
Matrix ans;
ans.row = a.row, ans.cal = b.cal;
ans = init(ans,0);
for(int i = 0; i < a.row; i++)
for(int j = 0; j < b.cal; j++)
for(int k = 0; k < a.cal; k++)
ans.m[i][j] = (ans.m[i][j] + a.m[i][k] * b.m[k][j])%mod;
return ans;
}
int quickpow(int a, int b, int mod)
{
int ans = 1;
for(;b;b >>= 1, a = a * 1ll * a % mod){
if(b & 1) ans = ans * 1ll * a % mod;
}
return ans;
}
int quick_pow(long long k, int b, Matrix A)
{
if(k < 0) return 0;
if(k == 0) return b;
Matrix I;
I.row = 3, I.cal = 1;
I = init(I, 0);
I.m[0][0] = 1;
I.m[1][0] = b;
I.m[2][0] = 0;
while(k){
if(k & 1) I = mul(A, I);
A = mul(A, A);
k>>=1;
}
return I.m[1][0]%mod;
}
int main (void)
{
int T;scanf("%d", &T);
while(T--){
int a, b, c, p;
long long n;
scanf("%I64d%d%d%d%d", &n, &a,&b, &c, &p);
if(a % p == 0){printf("0\n");continue;}
mod = p - 1;
Matrix A;
A.row = 3, A.cal = 3;
A = init(A, 0);
A.m[0][0] = A.m[2][1] = A.m[1][2] = 1;
A.m[1][0] = b;A.m[1][1] = c;
int res = quick_pow(n - 2, b, A);
printf("%d\n", quickpow(a, res, p));
}
}