HDU

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2631    Accepted Submission(s): 1186

Problem Description

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and i4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right. 

 

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231 as described above.

 

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

 

Sample Input

23 1 24 1 10

 

Sample Output

85369
Hint
In the first case, the third number is 85 = 2*1十2十3^4. In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <queue>#include <stack>#include <map>#include <cmath>#include <vector>#define max_ 200010#define inf 0x3f3f3f3f#define ll long longusing namespace std;struct mat{ll num[30][30];int n;}a;int n,m,x,y;mat mul(struct mat a,struct mat b){struct mat ans;ans.n=a.n;memset(ans.num,0,sizeof(ans.num));for(int i=1;i<=a.n;i++){for(int j=1;j<=a.n;j++){for(int k=1;k<=a.n;k++){ans.num[i][j]+=(a.num[i][k]*b.num[k][j])%2147493647UL;ans.num[i][j]%=2147493647UL;}}}return ans;}void show(struct mat a){printf("%d\n",a.n);for(int i=1;i<=a.n;i++){for(int j=1;j<=a.n;j++){printf("%d ",a.num[i][j]);}printf("\n" );}}int fpow(int k){struct mat ans,tmp=a;ans.n=a.n;memset(ans.num,0,sizeof(ans.num));for(int i=1;i<=a.n;i++){ans.num[i][i]=1;}while(k!=0){if(k&1)ans=mul(ans,tmp);tmp=mul(tmp,tmp);k/=2;}// show(ans);int d=(y*ans.num[1][1]+x*ans.num[1][2]+81*ans.num[1][3]+27*ans.num[1][4]+9*ans.num[1][5]+3*ans.num[1][6]+ans.num[1][7])%2147493647UL;return d;}int main(int argc, char const *argv[]) {a.n=7;memset(a.num,0,sizeof(a.num));a.num[1][1]=1;a.num[1][2]=2;a.num[1][3]=1;a.num[2][1]=1;a.num[3][3]=1;a.num[3][4]=4;a.num[3][5]=6;a.num[3][6]=4;a.num[3][7]=1;a.num[4][4]=1;a.num[4][5]=3;a.num[4][6]=3;a.num[4][7]=1;a.num[5][5]=1;a.num[5][6]=2;a.num[5][7]=1;a.num[6][6]=1;a.num[6][7]=1;a.num[7][7]=1;int t;cin>>t;while(t--){int k;cin>>k>>x>>y;if(k==1){printf("%d\n",x );continue;}else if(k==2){printf("%d\n",y );continue;}printf("%d\n",fpow(k-2));}return 0;}