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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9630 | Accepted: 6839 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626 6875
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
.
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
.
求斐波那契序列的公式。
由于该矩阵的特殊结构使得a(n+1)[0][0] = a(n)[0][0]+a(n)[0][1], a(n+1)[0][1] = a(n)[1][1], a(n+1)[1][0] = a(n)[0][1]+a(n)[1][0], a(n+1)[1][1] = a(n)[1][0];
code:
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<vector> 5 #include<algorithm> 6 #include<cmath> 7 #define M(a,b) memset(a,b,sizeof(a)) 8 9 using namespace std; 10 11 int n; 12 struct matrix 13 { 14 int a[2][2]; 15 void init() 16 { 17 a[0][0] = a[1][0] = a[0][1] = 1; 18 a[1][1] = 0; 19 } 20 }; 21 22 matrix mamul(matrix a,matrix b) 23 { 24 matrix c; 25 for(int i = 0;i<2;i++) 26 { 27 for(int j = 0;j<2;j++) 28 { 29 c.a[i][j] = 0; 30 for(int k = 0;k<2;k++) 31 c.a[i][j]+=(a.a[i][k]*b.a[k][j]); 32 c.a[i][j]%=10000; 33 } 34 } 35 return c; 36 } 37 38 matrix mul(matrix s, int k) 39 { 40 matrix ans; 41 ans.init(); 42 while(k>=1) 43 { 44 if(k&1) 45 ans = mamul(ans,s); 46 k = k>>1; 47 s = mamul(s,s); 48 } 49 return ans; 50 } 51 52 int main() 53 { 54 while(scanf("%d",&n)==1&n>=0) 55 { 56 if(n==0) puts("0"); 57 else 58 { 59 matrix ans; 60 ans.init(); 61 ans = mul(ans,n-1); 62 printf("%d\n",ans.a[0][1]%10000); 63 } 64 } 65 return 0; 66 }
下面代码只是测试公式，无法解决取模的问题，因为中间为double型，无法取模：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<vector> 5 #include<algorithm> 6 #include<cmath> 7 #define M(a,b) memset(a,b,sizeof(a)) 8 9 using namespace std; 10 11 double Pow(double a,int n) 12 { 13 double ans = 1; 14 while(n>=1) 15 { 16 if(n&1) 17 ans = a*ans; 18 n = n>>1; 19 a = a*a; 20 } 21 return ans; 22 } 23 24 int main() 25 { 26 int n; 27 double a = (sqrt(5.0)+1.0)/2; 28 double b = (-sqrt(5.0)+1.0)/2; 29 double c = (sqrt(5.0))/5; 30 while(scanf("%d",&n)==1) 31 { 32 int ans = (int)(c*(Pow(a,n)-Pow(b,n)))%10000; 33 printf("%d\n",ans); 34 } 35 return 0; 36 }
标签：des style blog http color io os ar for
原文地址：http://www.cnblogs.com/haohaooo/p/4019715.html
