Problem of Precision |
| Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 24 Accepted Submission(s): 20 |
| |
|
Problem Description
 |
|
Input
The first line of input gives the number of cases, T. T test
cases follow, each on a separate line. Each test case contains one
positive integer n. (1 <= n <= 10^9)
|
|
Output
For each input case, you should output the answer in one line.
|
|
Sample Input
3 1 2 5 |
|
Sample Output
9 97 841 |
共轭数？？
网上是这么说的
我只是说一下推理过程
(m+n)^k=an+根号6bn
(m-n)^k=an-根号6bn
我们把上面的k次方展开，那么只有奇数次方 （-n）^x(x属于奇数)才会对后面的产生影响，而影响加起来就是根号6bn (谢谢zhx～～)
1 /* 2 Author:wuhuajun 3 */ 4 #include <cmath> 5 #include <cstdio> 6 #include <algorithm> 7 #include <cstring> 8 #include <string> 9 #include <cstdlib> 10 using namespace std; 11 12 typedef long long ll; 13 typedef double dd; 14 const int maxn=210,mod=1024; 15 16 struct matrix 17 { 18 int m[3][3]; 19 } ans,base,c; 20 int n,T,sum; 21 22 void close() 23 { 24 25 exit(0); 26 } 27 28 matrix mul(matrix a,matrix b) 29 { 30 memset(c.m,0,sizeof(c.m)); 31 for (int i=1;i<=2;i++) 32 for (int j=1;j<=2;j++) 33 for (int k=1;k<=2;k++) 34 { 35 c.m[i][j]+=a.m[i][k]*b.m[k][j]; 36 c.m[i][j] %= mod; 37 } 38 return c; 39 } 40 41 void work() 42 { 43 memset(ans.m,0,sizeof(ans.m)); 44 ans.m[1][1]=ans.m[2][2]=1; 45 base.m[1][1]=5; 46 base.m[1][2]=12; 47 base.m[2][1]=2; 48 base.m[2][2]=5; 49 n--; 50 while (n!=0) 51 { 52 if (n & 1) 53 ans=mul(base,ans); 54 n/=2; 55 base=mul(base,base); 56 } 57 sum=ans.m[1][1]*5+ans.m[1][2]*2; 58 sum*=2; 59 sum--; 60 sum %= mod; 61 printf("%d ",sum); 62 } 63 64 65 void init() 66 { 67 scanf("%d",&T); 68 while (T--) 69 { 70 scanf("%d",&n); 71 work(); 72 } 73 } 74 75 int main () 76 { 77 init(); 78 close(); 79 return 0; 80 }