快速幂&矩阵快速幂

 快速幂模板

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main()
{
    ll a, b, c; cin >> a >> b >> c;
    ll aa = a, bb = b;
    ll ans = 1;
    while (b >= 1)
    {
        if (b & 1)
            ans = ans * a % c;
        a = a * a % c;
        b >>= 1;
    }
    printf("%lld^%lld mod %lld=%lld
", 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.
}

矩阵快速幂模板

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const long long maxn = 110;
long long mod = 1e9 + 7;
struct matrix
{
    long long v[maxn][maxn];
    long long n;
};
matrix mul(matrix a, matrix b)
{
    matrix res;
    res.n = a.n;
    memset(res.v, 0, sizeof(res.v));
    for (long long i = 1; i <= res.n; i++)
        for (long long j = 1; j <= res.n; j++)
            for (long long k = 1; k <= res.n; k++)
                res.v[i][j] += a.v[i][k] * b.v[k][j], res.v[i][j] %= mod;
    return res;
}
signed main()
{
    ll n, k; cin >> n >> k;
    matrix a, res;

    a.n = res.n = n;
    for (long long i = 1; i <= n; i++)
        for (long long j = 1; j <= n; j++)
            res.v[i][j] = i == j ? 1 : 0;
    for (long long i = 1; i <= n; i++)
        for (long long j = 1; j <= n; j++)
            cin >> a.v[i][j];
    while (k)
    {
        if (k & 1)
            res = mul(res, a);
        a = mul(a, a);
        k >>= 1;
    }
    for (long long i = 1; i <= n; i++)
        for (long long j = 1; j <= n; j++)
            printf("%lld%c", res.v[i][j], j == n ? '
' : ' ');
}