快速幂&矩阵快速幂
快速幂模板
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 5 int main() 6 { 7 ll a, b, c; cin >> a >> b >> c; 8 ll aa = a, bb = b; 9 ll ans = 1; 10 while (b >= 1) 11 { 12 if (b & 1) 13 ans = ans * a % c; 14 a = a * a % c; 15 b >>= 1; 16 } 17 printf("%lld^%lld mod %lld=%lld\n", aa, bb, c, ans % c);//注意：当b等于0的时候，结果为1，要单独对c取模，因为c可能为1. 18 }
矩阵快速幂模板
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const long long maxn = 110; 5 long long mod = 1e9 + 7; 6 struct matrix 7 { 8 long long v[maxn][maxn]; 9 long long n; 10 }; 11 matrix mul(matrix a, matrix b) 12 { 13 matrix res; 14 res.n = a.n; 15 memset(res.v, 0, sizeof(res.v)); 16 for (long long i = 1; i <= res.n; i++) 17 for (long long j = 1; j <= res.n; j++) 18 for (long long k = 1; k <= res.n; k++) 19 res.v[i][j] += a.v[i][k] * b.v[k][j], res.v[i][j] %= mod; 20 return res; 21 } 22 signed main() 23 { 24 ll n, k; cin >> n >> k; 25 matrix a, res; 26 27 a.n = res.n = n; 28 for (long long i = 1; i <= n; i++) 29 for (long long j = 1; j <= n; j++) 30 res.v[i][j] = i == j ? 1 : 0; 31 for (long long i = 1; i <= n; i++) 32 for (long long j = 1; j <= n; j++) 33 cin >> a.v[i][j]; 34 while (k) 35 { 36 if (k & 1) 37 res = mul(res, a); 38 a = mul(a, a); 39 k >>= 1; 40 } 41 for (long long i = 1; i <= n; i++) 42 for (long long j = 1; j <= n; j++) 43 printf("%lld%c", res.v[i][j], j == n ? '\n' : ' '); 44 }
