ZOJ 3707 Calculate Prime S

Description

Define S
 as the number of subsets of {1, 2, ..., n} that contain no consecutive integers. For each S
, if for all i (1 ≤ i < n) gcd(S[i],S
) is 1, we call that S
 as
a Prime S. Additionally, S[1] is also a Prime S. For the Kth minimum Prime S, we'd like to find the minimum S
 which is multiple of X and not less than the Kth minimum Prime
S. Please tell us the corresponding (S
 ÷ X) mod M.

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases.

For each test case, there are 3 integers K (1 ≤ K ≤ 106), X (3 ≤ X ≤ 100) and M (10 ≤ M ≤ 106), which are defined in above description.

Output

For each test case, please output the corresponding (S
 ÷ X) mod M.

Sample Input

1
1 3 10

Sample Output

1
数论题，没办法，本人数论比较渣，看人家题解搞懂的。
首先可以根据题目知道要找的数一定是斐波那契数，
然后，要知道斐波那契数的gcd是有结论的，详情点击。
然后剩下的就是矩阵乘法和余数的问题，余数这里同样有个巧妙的办法
x/y%z=x%(y*z)/y所以稍微转换一下就好了。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 2e7 + 10;
int T, n, m, x;
int p[maxn], f[maxn], tot;

struct martix
{
int a[2][2];
}a, b;

martix operator *(const martix &a, const martix &b)
{
martix c;
for (int i = 0; i < 2; i++)
{
for (int j = 0; j < 2; j++)
{
c.a[i][j] = 0;
for (int k = 0; k < 2; k++)
{
(c.a[i][j] += (LL)a.a[i][k] * b.a[k][j] % m) %= m;
}
}
}
return c;
}

void init()
{
p[tot++] = 4;
for (int i = 2; i < maxn; i++)
{
if (!f[i]) p[tot++] = i;
for (int j = 1; j < tot&&p[j] * i < maxn; j++)
{
f[p[j] * i] = 1;
if (i%p[j] == 0) break;
}
if (tot > 1e6) break;
}
for (int i = 0; i < 2; i++) swap(p[i], p[i + 1]);
}

martix get(martix x, int y)
{
martix c;
c.a[0][0] = c.a[1][1] = 1;
c.a[0][1] = c.a[1][0] = 0;
for (; y; y >>= 1)
{
if (y & 1) c = c*x;
x = x*x;
}
return c;
}

int main()
{
init();
scanf("%d", &T);
while (T--)
{
scanf("%d%d%d", &n, &x, &m);	m *= x;
a.a[0][0] = 1;	a.a[0][1] = 1;
b.a[0][0] = 0;	b.a[0][1] = 1;
b.a[1][0] = 1;	b.a[1][1] = 1;
a = a*get(b, p
- 2);
while (a.a[0][1] % x)
{
int y = a.a[0][0];
a.a[0][0] = a.a[0][1];
a.a[0][1] = (y + a.a[0][0]) % m;
}
printf("%d\n", a.a[0][1] / x);
}
return 0;
}