前缀和、差分、快速幂（取模）、矩阵快速幂

void Prefix_sum(const int a[], const int n) {//获得前缀和数组
  for (int i = 1;i <= n;++i) {
    Prefix__sum[i] = a[i] + Prefix__sum[i - 1];
  }
}

void Difference(const int a[], const int n) {//获得差分数组
  for (int i = 1;i <= n;++i) {
    Difference_[i] = a[i] - a[i - 1];
  }
  return;
}

void Interval_Maintenance(int l, int r, int k,int Difference_[]) {//对差分数组区间[l,r]上的所有数进行更新操作（k可能为负数）
  Difference_[l] += k;
  Difference_[r + 1] -= k;
  return;
  /*要想获得经过区间更新之后的数组中的第n个元素的值
  int sum =0;
  for (int i = 1;i <= n;++i) {
  sum += Difference_[i];
  }*/
}

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f3f3f3f3f
#define IOS ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
ll Quick_Pow_Mod(ll a, ll b, ll mod) {//求a的b次方
  ll ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans % mod * a % mod) % mod;
    }
    a = (a % mod * a % mod) % mod;
    b >>= 1;
  }
  return ans;
}
int main() {
  ll a, b, p;cin >> a >> b >> p;//求a的b次方模p
  cout << Quick_Pow_Mod(a, b,p) << endl;
}

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f3f3f3f3f
#define mod 1000000007
#define IOS ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
int n, m, p;
int a[maxn][maxn], b[maxn][maxn], c[maxn][maxn];
int main()
{
    cin >> n >> m >> p;//第一个矩阵为n行m列 第二个矩阵为m行p列 最终结果为n行p列的矩阵
    for (int i = 1;i <= n;++i) {
        for (int k = 1;k <= m;++k) {
            for (int j = 1;j <= p;++j) {
                c[i][j] += a[i][k] * b[k][j];
            }
        }
    }
    return 0;
}
```cpp

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f3f3f3f3f
#define mod 1000000007
#define IOS ios::sync_with_stdio(false)
#define endl '\n'
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 10;
ll n;
struct Matrix
{
    ll matrix[15][15];
    Matrix() {
        memset(matrix, 0, sizeof(matrix));//构造函数 创建Matrix类型变量时 自动调用
    }
};
Matrix mul(Matrix x, Matrix y)//矩阵乘法核心代码
{
    Matrix res;
    for (int i = 0;i < 2;i++)
        for (int k = 0;k < 2;k++)
            for (int j = 0;j < 2;j++)
                res.matrix[i][j] = (res.matrix[i][j] % mod + x.matrix[i][k] % mod * y.matrix[k][j] % mod) % mod;
    return res;
}
ll Quick_Matrix_Pow(ll p)
{
    Matrix base;
    Matrix ans;
    for (int i = 0;i < 2;i++)
        ans.matrix[i][i] = 1;//初始化答案矩阵（对角线上为1 ，其余为0）
    base.matrix[0][0] = base.matrix[0][1] = base.matrix[1][0] = 1;
    base.matrix[1][1] = 0;//初始化base矩阵
    while (p)//类似快速幂
    {
        if (1 & p) ans = mul(ans, base);//调用矩阵乘法
        base = mul(base, base);
        p >>= 1;
    }
    return ans.matrix[0][1];//ans.matrix = {f(n+1),f(n),f(n),f(n-1)}
                            //答案为ans.matrix[0][1]或ans.matrix[1][0]
}
int main()
{
    cin >> n;
    cout << Quick_Matrix_Pow(n) << endl;
    return 0;
}