BZOJ1002: [FJOI2007]轮状病毒

Description

给定n(N<=100)，编程计算有多少个不同的n轮状病毒。

Input

第一行有1个正整数n。

Output

将编程计算出的不同的n轮状病毒数输出

Sample Input

Sample Output

很遗憾，关于递推公式的印象已经变得模糊了。总而言之上马高精度，运算符重载不敢写，直接返回速度慢但我乐意，应该直接能看代码吧。

 1 #include <cstdio>
 2 #include <cstring>
 3 #define MAXN 1010
 4 using namespace std;
 5  
 6 struct BigInteger {
 7     int a[MAXN];
 8     const static int L = 1000;
 9      
10     BigInteger() {
11         memset(a, 0, sizeof(a)); 
12     }
13 };
14  
15 BigInteger f[1010][2];
16 int n;
17      
18     BigInteger Init(int N) {
19         BigInteger ans;
20         for (int i = 1; i <= ans.L; i++) {
21             ans.a[i] = N % 10;
22             N /= 10;
23             if (N == 0) break;
24         }
25         return ans;
26     }
27      
28     BigInteger Plus(BigInteger A, BigInteger B) {
29         BigInteger ans;
30         int k = 0;
31         for (int i = 1; i <= ans.L; i++) {
32             ans.a[i] = A.a[i] + B.a[i] + k;
33             k = ans.a[i] / 10, ans.a[i] %= 10;
34         }
35         return ans;
36     }
37      
38     BigInteger Minus(BigInteger A, BigInteger B) {
39         BigInteger ans;
40         for (int i = 1; i <= ans.L; i++) ans.a[i] = A.a[i];
41         for (int i = 1; i <= ans.L; i++) {
42             ans.a[i] -= B.a[i];
43             if (ans.a[i] < 0) { ans.a[i] += 10, ans.a[i + 1]--; }
44         }
45         return ans;
46     }
47      
48     BigInteger Times(BigInteger A, int B) {
49         BigInteger ans;
50         for (int i = 1; i <= ans.L; i++) {
51             ans.a[i] += A.a[i] * B;
52             if (ans.a[i] > 9) {
53                 ans.a[i + 1] += ans.a[i] / 10;
54                 ans.a[i] %= 10;
55             }
56         }
57         return ans;
58     }
59      
60     void Print(BigInteger A) {
61         int flag = 0;
62         for (int i = A.L; i >= 1; i--) {
63             if (flag) printf("%d", A.a[i]);
64             else if (A.a[i] > 0) {
65                 printf("%d", A.a[i]);
66                 flag = 1;
67             }
68         }
69         if (flag == 0) printf("0");
70         printf("
");
71     }
72      
73     void Prepare() {
74         f[0][1] = Init(1);
75         f[1][1] = Init(1), f[1][0] = Init(1);
76     }
77      
78 int main() {
79     scanf("%d", &n);
80     Prepare();
81     for (int i = 2; i < n; i++) {
82         f[i][0] = Plus(f[i - 1][0], f[i - 1][1]);
83         f[i][1] = Plus(f[i][0], f[i - 1][1]);
84     }
85     BigInteger ans;
86     for (int i = 1; i <= n; i++) ans = Plus(ans, Times(Times(f[n - i][1], i), i));
87     Print(ans);
88     return 0;
89 }

BZOJ1002