7.22 H Codeforces 392C Yet Another Number Sequence

题意：

给出n和k，表示一个序列A[i]，求前n项A[i]的和，结果% 1e9+7。
A[i] = i^k * F[i]，F[i] = F[i-1] + F[i-2]。

思路：

矩阵快速幂。

代码：

//coder:OX_louis#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>#include <ctime>#include <stack>using namespace std;typedef pair<int, int> pii;typedef long long ll;typedef unsigned long long ull;typedef vector<int> vi;#define MP(a,b) make_pair(a,b)#define pr(x) cout << #x << ": " << x << "  "#define pl(x) cout << #x << ": " << x << endl;#define pri(a) printf("%d\n",(a))#define xx first#define yy second#define sa(n) scanf("%d", &(n))#define sal(n) scanf("%lld", &(n))#define sai(n) scanf("%I64d", &(n))#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++)#define rep(i,a,b) for(int (i)=(a); (i)<(b); (i)++)#define per(i,a,b) for(int (i)=(b)-1; (i)>=(a); (i)--)#define CLR(a,i) memset((a),(i),sizeof(a))const int mod=10000;struct Mat{    int v[2][2];}mat[50],init;Mat mt(Mat a,Mat b){    Mat ans;    rep(i,0,2)    rep(j,0,2){        ans.v[i][j]=0;        rep(k,0,2){            ans.v[i][j]=(ans.v[i][j]+(a.v[k][j]*b.v[i][k])%mod)%mod;        }    }    return ans;}Mat k_pow(Mat a,int n){    Mat ans=init;    if(n==0)return ans;    int i=1;    while(n){        if(n&1){            ans=mt(ans,mat[i]);        }        i++;        n=n>>1;    }    return ans;}int main(){    mat[0].v[0][0]= 1 ;mat[0].v[0][1]= 0 ;    mat[0].v[1][0]= 0 ;mat[0].v[1][1]= 1 ;    int T,a,b,m,n;    sa(T);    while(T--){        sa(a); sa(b); sa(n); sa(m);        if(n==0) {pri(a);continue;}        else if(n==1) {pri(b);continue;}        mat[1].v[0][0]= 1 ;mat[1].v[0][1]= 1 ;        mat[1].v[1][0]= 1 ;mat[1].v[1][1]= 0 ;        init.v[0][0]= b ; init.v[0][1]= 0 ;        init.v[1][0]= a ; init.v[1][1]= 0 ;        rep(i,1,31){            mat[i+1]=mt(mat[i],mat[i]);        }        int mdd=1;        while(m--)mdd*=10;        Mat ans=k_pow(mat[1],n-1);        pri(ans.v[0][0]%mdd);    }    return 0;}