【递归】普通递归关系

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#include <ext/rope>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define vi vector<int>
#define pii pair<int,int>
#define mod 1000000007
#define inf 0x3f3f3f3f
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
const int maxn=3e5+10;
const int dis[][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
using namespace __gnu_cxx;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
struct mat
{
    int n,m,a[2][2];
    mat(int i=0,int j=0)
    {
        n=i,m=j;
        memset(a,0,sizeof(a));
    }
    mat operator*(const mat&p)const
    {
        mat q(n,p.m);
        int i,j,k;
        rep(i,0,n-1)rep(j,0,m-1)rep(k,0,p.m-1)
        q.a[i][k]=(q.a[i][k]+1LL*a[i][j]*p.a[j][k])%mod;
        return q;
    }
};
int main()
{
    int i,j,m,k,t;
    mat f(1,2);
    ll a,b,n,x;
    scanf("%lld%lld%lld%lld%lld",&f.a[0][0],&f.a[0][1],&a,&b,&n);
    mat g(2,2);
    g.a[0][0]=0,g.a[0][1]=b,g.a[1][0]=1,g.a[1][1]=a;
    if(n==0)exit(0*printf("%d\n",f.a[0][0]));
    n--;
    while(n)
    {
        if(n&1)f=f*g;
        g=g*g;
        n>>=1;
    }
    printf("%d\n",f.a[0][1]);
    //system("pause");
    return 0;
}