| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7715 | Accepted: 5474 |
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:
.
Source
1 #include<iostream> 2 #include<cstdio> 3 #include<cstdlib> 4 #include<cstring> 5 6 struct node 7 { 8 int a[5][5]; 9 }now,cur; 10 11 12 void make_first() 13 { 14 now.a[1][1]=1; 15 now.a[1][2]=1; 16 now.a[2][1]=1; 17 now.a[2][2]=0; 18 19 cur.a[1][1]=1; 20 cur.a[1][2]=1; 21 cur.a[2][1]=1; 22 cur.a[2][2]=0; 23 } 24 25 struct node make_cheng(node a,node b) //发现结构体忘记了。 26 { 27 struct node f; 28 int i,k,j; 29 memset(f.a,0,sizeof(f.a)); 30 for(i=1;i<=2;i++) 31 for(k=1;k<=2;k++) 32 if(a.a[i][k]) 33 { 34 for(j=1;j<=2;j++) 35 if(b.a[k][j]) 36 { 37 f.a[i][j]+=a.a[i][k]*b.a[k][j]; 38 if(f.a[i][j]>=10000) 39 f.a[i][j]%=10000; 40 } 41 } 42 return f; 43 } 44 45 void make_EF(int n) 46 { 47 make_first(); 48 while(n) 49 { 50 if(n&1) 51 { 52 now=make_cheng(now,cur);//！！！ 53 } 54 n=n/2; 55 cur=make_cheng(cur,cur); 56 } 57 printf("%d ",now.a[1][1]); 58 } 59 60 int main() 61 { 62 int n; 63 while(scanf("%d",&n)>0) 64 { 65 if(n==-1)break; 66 if(n==0){printf("0 ");continue;} 67 if(n==1||n==2){printf("1 ");continue;} 68 make_EF(n-2); 69 } 70 return 0; 71 }