[杂题]CSUOJ1413 Area of a Fractal

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <climits>
 5 #include <cctype>
 6 #include <cmath>
 7 #include <string>
 8 #include <sstream>
 9 #include <iostream>
10 #include <algorithm>
11 #include <iomanip>
12 using namespace std;
13 #include <queue>
14 #include <stack>
15 #include <vector>
16 #include <deque>
17 #include <set>
18 #include <map>
19 typedef long long LL;
20 typedef long double LD;
21 const double pi=acos(-1.0);
22 const double eps=1e-6;
23 #define INF 0x3f3f3f
24 #define lson l, m, rt<<1
25 #define rson m+1, r, rt<<1|1
26 typedef pair<int, int> PI;
27 typedef pair<int, PI > PP;
28 #ifdef _WIN32
29 #define LLD "%I64d"
30 #else
31 #define LLD "%lld"
32 #endif
33 //#pragma comment(linker, "/STACK:1024000000,1024000000")
34 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
35 //inline int read(){char ch=‘ ‘;int ans=0;while(ch<‘0‘ || ch>‘9‘)ch=getchar();while(ch<=‘9‘ && ch>=‘0‘){ans=ans*10+ch-‘0‘;ch=getchar();}return ans;}
36 inline void print(LL x){printf(LLD, x);puts("");}
37 //inline void read(LL &ret){char c;int sgn;LL bit=0.1;if(c=getchar(),c==EOF) return ;while(c!=‘-‘&&c!=‘.‘&&(c<‘0‘||c>‘9‘)) c=getchar();sgn=(c==‘-‘)?-1:1;ret=(c==‘-‘)?0:(c-‘0‘);while(c=getchar(),c>=‘0‘&&c<=‘9‘) ret=ret*10+(c-‘0‘);if(c==‘ ‘||c==‘\n‘){ ret*=sgn; return ; }while(c=getchar(),c>=‘0‘&&c<=‘9‘) ret+=(c-‘0‘)*bit,bit/=10;ret*=sgn;}
38 const int mod=1000000007;
39 
40 LL quick(LL a, LL b)
41 {
42     LL ans=1;
43     while(b)
44     {
45         if(b & 1)ans=(ans*a)%mod;
46         a=(a*a)%mod;
47         b>>=1;
48     }
49     return ans%mod;
50 }
51 
52 void ex_gcd(int a, int b, int &x, int &y)
53 {
54     if(b)
55     {
56         ex_gcd(b, a%b, x, y);
57         int tmp=x;
58         x=y;
59         y=tmp-(a/b)*y;
60     }
61     else
62     {
63         x=1, y=0;
64         return ;
65     }
66 }
67 int main()
68 {
69     int t;
70     scanf("%d", &t);
71     while(t--)
72     {
73         int n;
74         scanf("%d", &n);
75         if(n==0)
76         {
77             printf("1\n");
78             continue;
79         }
80         int x, y;
81         ex_gcd(5, mod, x, y);
82         print((((3*quick(17, n))%mod+(2*quick(7, n))%mod)*x)%mod);
83     }
84     return 0;
85 }