快速幂和矩阵快速幂模板

Posted 2020-10-08 ╰追憶似水年華ぃ╮

快速幂模板：

ll qmod(ll x,ll n,ll mod)
{
    ll res=1;
    while(n){
        if(n&1) res=(res*x)%mod;
        x=(x*x)%mod;
        n/=2;
    }
    return res;
}

例题：hdu 1097

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const ll mod=10;

ll qmod(ll x,ll n)
{
    ll res=1;
    while(n){
        if(n&1) res=(res*x)%mod;
        x=(x*x)%mod;
        n/=2;
    }
    return res;
}

int main()
{
    ll a,b;
    while(cin>>a>>b)
    {
        cout<<qmod(a,b)<<endl;
    }
    return 0;
}

hdu 1097

 

矩阵快速幂：

 1 mat mul(mat &A, mat &B)
 2 {
 3     mat C(A.size(), vec(B[0].size()));///分配大小，A的行，B的列
 4     for (int i = 0; i<A.size(); i++) ///矩阵A的行
 5         for (int k = 0; k<B.size(); k++) ///矩阵B的行
 6             for (int j = 0; j<B[0].size(); j++) ///矩阵B的列
 7                 C[i][j] = (C[i][j] + A[i][k] * B[k][j] % mod + mod) % mod;
 8     return C;
 9 }
10 ///计算A^n
11 mat pow(mat A, LL n)
12 {
13     mat B(A.size(), vec(A.size()));///和矩阵A的大小相同
14     for (int i = 0; i<A.size(); i++)
15         B[i][i] = 1;
16     while (n>0)
17     {
18         if (n & 1) B = mul(B, A);
19         A = mul(A, A);
20         n >>= 1;
21     }
22     return B;
23 }
24 void solve()
25 {
26     mat A(2, vec(2));///2*2的矩阵
27     A[0][0] = 1;
28     A[0][1] = 1;
29     A[1][0] = 1;
30     A[1][1] = 0;
31     A = pow(A, n);
32     printf("%d\\n", (A[1][0] % mod - 1 + mod) % mod);
33 }

例题：求F(2n+3)-1

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <vector>
 5 using namespace std;
 6 const int mod = 998244353;
 7 typedef long long LL;
 8 LL n;
 9 typedef vector<LL>vec;
10 typedef vector<vec>mat;
11 mat mul(mat &A, mat &B)
12 {
13     mat C(A.size(), vec(B[0].size()));///分配大小，A的行，B的列
14     for (int i = 0; i<A.size(); i++) ///矩阵A的行
15         for (int k = 0; k<B.size(); k++) ///矩阵B的行
16             for (int j = 0; j<B[0].size(); j++) ///矩阵B的列
17                 C[i][j] = (C[i][j] + A[i][k] * B[k][j] % mod + mod) % mod;
18     return C;
19 }
20 ///计算A^n
21 mat pow(mat A, LL n)
22 {
23     mat B(A.size(), vec(A.size()));///和矩阵A的大小相同
24     for (int i = 0; i<A.size(); i++)
25         B[i][i] = 1;
26     while (n>0)
27     {
28         if (n & 1) B = mul(B, A);
29         A = mul(A, A);
30         n >>= 1;
31     }
32     return B;
33 }
34 void solve()
35 {
36     mat A(2, vec(2));///2*2的矩阵
37     A[0][0] = 1;
38     A[0][1] = 1;
39     A[1][0] = 1;
40     A[1][1] = 0;
41     A = pow(A, n);
42     printf("%d\\n", (A[1][0] % mod - 1 + mod) % mod);
43 }
44 int main()
45 {
46     while (~scanf("%lld", &n))
47     {
48         n = 2 * n + 3;
49         solve();
50     }
51 }

http://www.cnblogs.com/zxhyxiao/p/7577282.html