hdu2481

先算出总的状态转移方程：dp[i]  = 3*dp[i-1]-dp[i-2]+2; 与bzoj1002题转移方程相同。

因为n对于m可能无逆元，所以mod = n*m。可能会爆longlong。所以用类比于快速幂的写法，写快速乘。然后用矩阵快速幂+欧拉函数优化。

Toy

Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 710    Accepted Submission(s): 377

Problem Description

On birthday, Anthony got a toy. It is constructed with N+1(N>=3) balls and 2*N sticks. All balls are in a same plane. One of them is special, while the other N balls are connected to it by N sticks with the same length. The angles between any two adjacent sticks are equal. And finally, any two adjacent balls(except the central one) are connected by a stick.
  Here are two examples:

Anthony wanted to remove N sticks, leaving all balls still connected. He wanted to know the number of all legal solutions. Your task is to solve this problem for him.
  Notice that if a solution will be the same as another one by rotation, these two solutions should be consider as the same.
The answer may be quite large. You just need to calculate the remainder of the answer when divided by M.

 

Input

Input contains several test cases.
For each test case, there is only one line containing two integers N and M(3<=N<=10^9, 2<=M<=10^9).
Input is terminated by EOF.

 

Output

For each case, output one integer in one line, representing the remainder of the number of all solutions when divided by M.

 

Sample Input

3 100004 100004 10

 

Sample Output

6133

#include"bits/stdc++.h"using namespace std;typedef long long LL;const int maxn = 1e5 + 7;LL mod;bool is_prime[maxn];int prime[maxn], tail;struct matrix {    LL m[3][3];};LL A[3][3] = {{3, -1, 1}, {1, 0, 0}, {0, 0, 1}};LL qmul(LL a, LL b) {    LL ret = 0;    a %= mod;    b %= mod;    if(a < 0) a += mod;    if(b < 0) b += mod;    while(b) {        if(b & 1) ret = (ret + a) % mod;        a = (a+a) % mod;        b >>= 1;    }    return ret;}void mul(matrix &A, matrix &B) {    matrix C;    memset(C.m, 0, sizeof(C.m));    for(int i = 0; i < 3; i++)        for(int j = 0; j < 3; j++)            for(int k = 0; k < 3; k++)                C.m[i][j] = (C.m[i][j] + qmul(A.m[i][k],B.m[k][j])) % mod;    for(int i = 0; i < 3; i++)        for(int j = 0; j < 3; j++)            A.m[i][j] = C.m[i][j];}matrix qpow(matrix a, LL n) {    matrix ret;    for(int i = 0; i < 3; i++)        for(int j = 0; j < 3; j++)            ret.m[i][j] = (i == j);    while(n) {        if(n & 1) mul(ret, a);        mul(a, a);        n >>= 1;    }    return ret;}void getprime() {    int n = 32000;    for(int i = 2; i <= n; i++)        is_prime[i] = 1;    for(int i = 2; i <= n; i++) {        if(is_prime[i]) {            prime[++tail] = i;            for(int j = 2 * i; j <= n; j += i) {                is_prime[j] = 0;            }        }    }}LL phi(int n) {    int ret = n;    for(int i = 1; prime[i]*prime[i] <= n; i++) {        if(n % prime[i] == 0) {            ret -= ret / prime[i];            while(n % prime[i] == 0) n /= prime[i];        }    }    if(n > 1) ret -= ret / n;    return ret;}LL f(LL n) {    if(n == 1) return 1;    if(n == 2) return 5;    matrix temp;    for(int i = 0; i < 3; i++)        for(int j = 0; j < 3; j++)            temp.m[i][j] = A[i][j];    matrix ret = qpow(temp, n - 2);    LL ans = (qmul(5, ret.m[0][0]) + ret.m[0][1] + 2*ret.m[0][2]) % mod;    return ans;}int main() {#ifdef ___LOCAL_WONZY___    freopen ("input.txt", "r", stdin);#endif // ___LOCAL_WONZY___    int n, m;    LL ans;    getprime();    while(~scanf("%d%d", &n, &m)) {        ans = 0;        mod = (LL) m * n;        int i;        for(i = 1; i * i < n; i++) {            if(n % i == 0) {                ans = (ans + qmul(phi(i), f(n / i)) ) % mod;                ans = (ans + qmul(phi(n / i), f(i)) ) % mod;            }        }        if(i * i == n) ans = (ans + qmul(phi(i), f(i))) % mod;        cout << (ans/n) % (mod / n) << endl;    }    return 0;}