bzoj 1089 [SCOI2003]严格n元树（DP+高精度）

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream> 
 4 using namespace std;
 5 
 6 const int maxn = 5000+10;
 7 const int rad = 1000;
 8 
 9 struct Bign {  int N[maxn],len;  };
10 void print(Bign a) {
11     printf("%d",a.N[a.len-1]);
12     for(int i=a.len-2;i>=0;i--)
13         printf("%03d",a.N[i]);            //补全位数 
14     putchar(‘\n‘);
15 }
16 Bign operator *(Bign A,Bign B) {
17     Bign C;
18     int lena=A.len,lenb=B.len;
19     for(int i=0;i<lena+lenb;i++) C.N[i]=0;
20     for(int i=0;i<lena;i++)
21         for(int j=0;j<lenb;j++)
22             C.N[i+j] += A.N[i]*B.N[j];
23     C.len=A.len+B.len;
24     for(int i=0;i<C.len;i++) 
25         if(C.N[i]>=rad) {
26             if(i==C.len-1) 
27                 C.len++ , C.N[i+1]=C.N[i]/rad;
28             else C.N[i+1]+=C.N[i]/rad;
29             C.N[i]%=rad;
30         }
31     while(C.len && !C.N[C.len-1]) C.len--;
32     return C;
33 }
34 Bign operator ^(Bign A,int p) {            //快速幂 
35     Bign C;
36     C.len=1; C.N[0]=1;
37     while(p) {
38         if(p&1) C=C*A; A=A*A;  p>>=1;
39     }
40     return C;
41 }
42 Bign operator +(Bign A,int x) {
43     A.N[0]+=x;
44     int now=0;
45     while(A.N[now]>=rad) {
46         A.len=max(A.len,now+1);
47         A.N[now+1]+=A.N[now]/rad;
48         A.N[now]%=rad;
49         now++;
50         A.len=max(A.len,now);
51     }
52     return A;
53 }
54 Bign operator-(Bign A,Bign B) {
55     for(int i=0;i<A.len;i++) {
56         A.N[i]-=B.N[i];
57         if(A.N[i]<0)
58             A.N[i]+=rad , A.N[i+1]--;
59     }
60     while(A.len && !A.N[A.len-1]) A.len--;
61     return A;
62 }
63 
64 int n,d;
65 Bign S[50];
66 
67 int main() {
68     scanf("%d%d",&n,&d);
69     if(!d) { puts("1"); return 0; }
70     S[0].len=1; S[0].N[0]=1;
71     for(int i=1;i<=d;i++)
72         S[i]=(S[i-1]^n)+1;
73     print(S[d]-S[d-1]);
74     return 0;
75 }