HDU

给你下面程序：

#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include <cstdio> 
#include<iostream> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
#include<vector> 

const int MAX=100000*2; 
const int INF=1e9; 

int main() 
{ 
  int n,m,ans,i; 
  while(scanf("%d%d",&n,&m)!=EOF) 
  { 
    ans=0; 
    for(i=1;i<=n;i++) 
    { 
      if(i&1)ans=(ans*2+1)%m; 
      else ans=ans*2%m; 
    } 
    printf("%d
",ans); 
  } 
  return 0; 
}

---

【数据范围】
1≤n,m≤1000000000$1\le n,m\le 1000000000$

---

【分析】
通过打表得知fi=fi−2+2i−1$f_i=f_{i-2}+2^{i-1}$,因为n太大肯定不能暴力做，开始想着通过这个公式等比数列求和就可以做，但是后面发现因为要Mod m$Modm$，而m的值不是定值，求逆元就不一定有。所以构造矩阵来做：

⎡⎣⎢fnfn−12i⎤⎦⎥=⎡⎣⎢010100102⎤⎦⎥×⎡⎣⎢fn−1fn−22i−1⎤⎦⎥$\left[\begin{array}{c}{f}_n\\ f_{n-1}\\ 2^i\end{array}\right]=\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 0\\ 0& 0& 2\end{array}\right]\times \left[\begin{array}{c}{f}_{n-1}\\ f_{n-2}\\ 2^{i-1}\end{array}\right]$
由此可转移如下：
⎡⎣⎢fnfn−12i⎤⎦⎥=⎡⎣⎢010100102⎤⎦⎥n−2×⎡⎣⎢f2f122⎤⎦⎥(n≥3)$\left[\begin{array}{c}{f}_n\\ f_{n-1}\\ 2^i\end{array}\right]={\left[\begin{array}{ccc}0& 1& 1\\ 1& 0& 0\\ 0& 0& 2\end{array}\right]}^{n-2}\times \left[\begin{array}{c}{f}_2\\ f_1\\ 2^2\end{array}\right](n\ge 3)$

fn=n(n≤2)$f_n=n(n\le 2)$

---

【代码】

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long LL;
typedef vector<LL> vec;
typedef vector<vec> mat;
void Init(mat &A) {//构造矩阵
    A[0][0] = 0; A[0][1] = 1LL, A[0][2] = 1LL;
    A[1][0] = 1LL; A[1][1] = 0, A[1][2] = 0;
    A[2][0] = 0; A[2][1] = 0, A[2][2] = 2LL;
}
mat mul(mat &A, mat &B, LL m) {//矩阵乘法
    mat C(A.size(), vec(B[0].size()));
    for(int i = 0; i < A.size(); i++)
        for(int k = 0; k < B.size(); k++)
            for(int j = 0; j < B[0].size(); j++)
                C[i][j] = (C[i][j] + A[i][k] * B[k][j] % m) % m;
    return C;
}
mat pow(mat &A, LL n, LL m) {//矩阵快速幂
    mat B(A.size(), vec(A.size()));
    for(int i = 0; i < A.size(); i++)B[i][i] = 1LL;
    while(n > 0) {
        if(n & 1)B = mul(B, A, m);
        n >>= 1;
        A = mul(A, A, m);
    }
    return B;
}
void solve(mat A, LL n, LL m) {
    LL res = 0;
    if(n <= 2LL)res = n % m;
    else {
        A = pow(A, n - 2LL, m);
        res = (A[0][0] * 2LL + A[0][1] + A[0][2] * 4LL % m) % m;
    }
    printf("%lld
", res);
}
int main() {
    LL n, m;
    while(~scanf("%lld%lld", &n, &m)) {
        mat A(3, vec(3));
        Init(A);
        solve(A, n, m);
    }
    return 0;
}