【费马小定理+快速幂+逆元】BZOJ3240-[NOI2013]矩阵游戏

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define MOD 1000000007
 6 using namespace std;
 7 typedef long long ll;
 8 const int MAXN=1000000+5;
 9 struct node
10 {
11     ll uni,ord;//uni表示a≠1，ord表示a=1 
12 }n,m;
13 ll a,b,c,d;
14 
15 ll quick(ll x,ll p)
16 {
17     ll res=p,retu=1,now=x;
18     while (res>0)
19     {
20         if (res&1) retu=(retu*now) % MOD;
21         now=(now*now) % MOD;
22         res>>=1;
23     }
24     return retu;
25 }
26 
27 ll ni(ll x)
28 {
29     return quick(x,MOD-2);
30 }
31 
32 node get_value(char str[])
33 {
34     int len=strlen(str);
35     node ret=(node){0,0}; 
36     for (int i=0;i<len;i++)
37     {
38         ret.uni=((ret.uni*10)%(MOD-1)+str[i]-‘0‘)%(MOD-1);
39         ret.ord=((ret.ord*10)%MOD+str[i]-‘0‘)%MOD;
40     }
41     return ret;
42 } 
43 
44 void init()
45 {
46     char strm[MAXN],strn[MAXN];
47     scanf("%s%s",strn,strm);
48     n=get_value(strn);
49     m=get_value(strm);
50     scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
51 }
52 
53 void get_ans()
54 {
55     ll f;//=f[n+1][1]
56     if (a==1)
57     {
58         ll D=((((c*(m.ord-1))%MOD)*b)%MOD+d)%MOD;
59         if (c==1) f=(1+n.ord*D)%MOD;
60             else
61             {
62                 ll cn=quick(c,n.uni);
63                 f=(cn+(D*(cn-1)*ni(c-1)%MOD))%MOD; 
64             } 
65     }
66     if (a!=1)
67     {
68         ll am=quick(a,m.uni-1);
69         ll A=(am*c)%MOD;
70         ll B=(((((b*c)%MOD*(am-1))%MOD*ni(a-1)))%MOD+d)%MOD;
71         ll An=quick(A,n.uni);
72         f=An+((B*(An-1)%MOD)*ni(A-1))%MOD;
73     }
74     printf("%lld",((f-d)*ni(c)%MOD+MOD)%MOD); 
75 }
76 
77 int main()
78 {
79     init();
80     get_ans();
81     return 0;
82 }