HDU 1568

Posted 2020-09-30 xFANx

给定n，0 <= n <= 100000000，斐波那契数列的第n项的前四位（不足则全部输出）

 

利用不动点去解得斐波那契数列的公式，显然数据爆longlong，所以以10为底取对数即可

 

前20项不足10000，预处理会更方便

 

 1 #include <cmath>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <queue>
 6 #include <vector>
 7 #define max(x, y) (x > y ? x : y)
 8 #define min(x, y) (x > y ? y : x)
 9 #define INF 0x3f3f3f3f
10 #define mod 1000000007
11 #define Yes printf("Yes\n")
12 #define No printf("No\n")
13 using namespace std;
14 typedef long long LL;
15 typedef pair<int, int> PII;
16 
17 int n;
18 
19 const int maxn = 30;
20 
21 int f[maxn] = {0, 1, 1};
22 
23 int main(int argc, const char * argv[]) {
24     for (int i = 3; i <= 20; i++) {
25         f[i] = f[i - 1] + f[i - 2];
26     }
27     //printf("%d\n", f[20]);
28     while (~scanf("%d", &n)) {
29         if (n <= 20) {
30             printf("%d\n", f[n]);
31         } else {
32             double tmp = -0.5 * log(5.0) / log(10.0) + ((double)n) * log((sqrt(5.0)+1.0)/2.0) / log(10.0);
33             tmp -= floor(tmp);
34             tmp = pow(10.0, tmp);
35             while (tmp < 1000) tmp *= 10;
36             printf("%d\n", (int)(tmp));
37         }
38     }
39     return 0;
40 }