HDU 2256 構造矩陣

Problem of Precision

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 289    Accepted Submission(s): 151

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

/* 構造過程： (sqrt(2)+sqrt(3))^2=5+2*sqrt(6); (5+2*sqrt(6))^n=An+Bn*sqrt(6); (5-2*sqrt(6))^n=An-Bn*sqrt(6); //因爲此項等於(0.101...)^n 約等於0 An+Bn*sqrt(6)=(A(n-1)+B(n-1)*sqrt(6))*(5+2*sqrt(6))=(5*A(n-1)+12B(n-1))+(5*B(n-1)+2*A(n-1))*sqrt(6) 所以有： An=5*A(n-1)+12B(n-1) Bn=2*A(n-1)+5*B(n-1) (5+2*sqrt(6))^n=An+Bn*sqrt(6)+(5-2*sqrt(6))^n=An-Bn*sqrt(6)=2An-(0.101...)^n 所以(5-2*sqrt(6))^n=2An-1 最後的答案ans=2An-1; 而矩陣爲 | 5 12| | 2 5| */ #include<stdio.h> struct node { long long num[2][2]; }; node res,tp; node mut(node a,node b) { node c; int i,j,k; for(i=0;i<2;i++) for(j=0;j<2;j++) { c.num[i][j]=0; for(k=0;k<2;k++) c.num[i][j]+=(a.num[i][k]*b.num[k][j])%1024; c.num[i][j]%=1024; } return c; } node abc(int temp,node p) { tp=p; p=res; while(temp) { if(temp&1) { p=mut(p,tp); temp--; } else { tp=mut(tp,tp); temp>>=1; } } return p; } int main() { node p; res.num[0][0]=res.num[1][1]=1; res.num[0][1]=res.num[1][0]=0; p.num[0][0]=p.num[1][1]=5; p.num[0][1]=12;p.num[1][0]=2; int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); if(n==1) { printf("9/n"); continue; } node k=abc(n-1,p); int sum=(k.num[0][0]*5+k.num[0][1]*2)%1024; sum=sum*2-1; sum=sum%1024; printf("%d/n",sum); } return 0; }