HDU --- 4686 【矩阵快速幂+求和】

传送门

思路: 这道题也是给了公式, 然后跟着公式推导, 也很容易的知道矩阵的构造, 注意的是在矩阵中把Sn加上, 这样就可以直接算出答案了.

AC Code

#include<cstdio>#include<cmath>#include<algorithm>#include<iostream>#include<cstring>#include<set>#include<queue>#include<functional>#include<vector>#include<stack>#include<map>#include<cstdlib>#define CLR(x) memset(x,0,sizeof(x))#define ll long long int#define PI acos(-1.0)#define db double#define mod 1000000007using namespace std;const int maxn=1e5+5;const db eps=1e-6;const int inf=1e9;const ll INF=1e15;int cas=1;int ssize;struct Ma{    ll a[10][10];    void cc()    {        memset(a,0,sizeof(a));    }    Ma operator * (const Ma & b) const {        Ma tmp;        tmp.cc();        for(int i=0;i<ssize;i++){            for(int j=0;j<ssize;j++){                for(int k=0;k<ssize;k++){                    tmp.a[i][j] += (a[i][k] * b.a[k][j]);                    tmp.a[i][j] %= mod;                }            }        }        return tmp;    }}res,x;ll A0,AX,AY,B0,BX,BY;void init(){    ssize = 5;    res.cc();    x.cc();    for(int i=0;i<ssize;i++)        res.a[i][i] = 1 ;    x.a[0][0] = 1;    x.a[1][0] = x.a[1][1] = AX*BX%mod;    x.a[2][0] = x.a[2][1] = AX*BY%mod;  x.a[2][2] = AX%mod;    x.a[3][0] = x.a[3][1] = BX*AY%mod;  x.a[3][3] = BX%mod;    x.a[4][0] = x.a[4][1] = AY*BY%mod;  x.a[4][2] = AY%mod; x.a[4][3] = BY%mod;    x.a[4][4] = 1;        //构造矩阵. }void qpow(ll n){    while(n)    {         //printf("%lld\n",n);        if(n&1) res = res * x;        x = x*x ;        n /= 2;    }}int main(){    ll n;    ll b[10];    while(~scanf("%lld",&n)){        scanf("%lld%lld%lld%lld%lld%lld",&A0,&AX,&AY,&B0,&BX,&BY);        if(n == 0){       //注意特判等于0的时候, 否则会T.            printf("0\n");            continue;        }        b[0] = b[1] = A0*B0%mod;        b[2] = A0%mod;        b[3] = B0%mod;        b[4] = 1;        init();        qpow(n-1);        ll ans = 0;        for(int i=0;i<ssize;i++){            ans += (res.a[i][0] * b[i]);            ans %= mod;        }        printf("%lld\n",ans);    }}