loj2253 「SNOI2017」礼物

对于一个在位置 \(i\) 的数，他等于 \(i^k+sum_{1,k-1}\)。
二项式定理推 \(i^k\)，矩阵快速幂即可。

#include 
#include 
using namespace std;
typedef long long ll;
int k, c[15][15];
ll n;
const int mod=1e9+7;
struct Matrix{
    int num[15][15];
    Matrix operator*(const Matrix &x)const{
        Matrix re;
        for(int i=0; i<=k+1; i++)
            for(int j=0; j<=k+1; j++){
                re.num[i][j] = 0;
                for(int l=0; l<=k+1; l++)
                    re.num[i][j] = (re.num[i][j] + (ll)num[i][l]*x.num[l][j]) % mod;
            }
        return re;
    }
}yua, zhu, dan;
Matrix ksm(ll b){
    while(b){
        if(b&1) dan = dan * zhu;
        zhu = zhu * zhu;
        b >>= 1;
    }
    return dan;
}
int main(){
    cin>>n>>k;
    for(int i=0; i<=k; i++)
        yua.num[0][i] = dan.num[i][i] = 1;
    c[0][0] = zhu.num[k][k+1] = dan.num[k+1][k+1] = 1;
    for(int i=1; i<=k; i++){
        c[i][0] = 1;
        for(int j=1; j<=i; j++)
            c[i][j] = (c[i-1][j] + c[i-1][j-1]) % mod;
    }
    for(int i=0; i<=k; i++)
        for(int j=i; j<=k; j++)
            zhu.num[i][j] = c[j][i];
    zhu.num[k+1][k+1] = 2;
    Matrix ans=yua*ksm(n-1);
    cout<<(ans.num[0][k+1]+ans.num[0][k])%mod<

转载于:https://www.cnblogs.com/poorpool/p/8571759.html