hdu 5950 数学公式 + 矩阵快速幂

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2205    Accepted Submission(s): 976

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, andi4. 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.解： 将n^4 化简 n^4=(n-1)^4+4*(n-1)^3+6*(n-1)^2+4*(n-1)^+1;具体构造矩阵请看 代码：
#include <iostream>#include <stdio.h>#include <string.h>#define N 7using namespace std;typedef long long ll;const ll mod=2147493647;ll a[N][N],temp[N][N],res[N][N],c[3];ll x,y,n;void init(){   a[0][0]=1,a[0][1]=2,a[0][2]=1,a[0][3]=4,a[0][4]=6,a[0][5]=4,a[0][6]=1;   a[1][0]=1,a[1][1]=0,a[1][2]=0,a[1][3]=0,a[1][4]=0,a[1][5]=0,a[1][6]=0;   a[2][0]=0,a[2][1]=0,a[2][2]=1,a[2][3]=4,a[2][4]=6,a[2][5]=4,a[2][6]=1;   a[3][0]=0,a[3][1]=0,a[3][2]=0,a[3][3]=1,a[3][4]=3,a[3][5]=3,a[3][6]=1;   a[4][0]=0,a[4][1]=0,a[4][2]=0,a[4][3]=0,a[4][4]=1,a[4][5]=2,a[4][6]=1;   a[5][0]=0,a[5][1]=0,a[5][2]=0,a[5][3]=0,a[5][4]=0,a[5][5]=1,a[5][6]=1;   a[6][0]=0,a[6][1]=0,a[6][2]=0,a[6][3]=0,a[6][4]=0,a[6][5]=0,a[6][6]=1;}void pw(ll a[N][N],ll b[N][N]){   memset(temp,0,sizeof(temp));    for (int i=0;i<N;i++)       for (int j=0;j<N;j++)         for (int k=0;k<N;k++)            temp[i][j]=(temp[i][j]%mod+a[i][k]*b[k][j]%mod)%mod;    for (int i=0;i<N;i++)       for (int j=0;j<N;j++)            a[i][j]=temp[i][j];}void ppw(ll a[N][N],ll nu){    memset(res,0,sizeof(res));     for (int i=0;i<N;i++)      res[i][i]=1;     while (nu)     {        if (nu&1)        pw(res,a);        pw(a,a);        nu>>=1;     }}void init1(){   c[0]=(2*x%mod+y%mod+81)%mod;   c[1]=y%mod;}int main(){    int t;    ll ans;    cin>>t;    while (t--)    {      scanf("%lld%lld%lld",&n,&x,&y);      ans=0;      if (n==1)      cout<<x%mod<<endl;      else if (n==2)      cout<<y%mod<<endl;      else if (n==3)      cout<<(2*x%mod+y%mod+81)%mod<<endl;      else      {         init();         init1();         ppw(a,n-3);         ans=(ans+c[0]*res[0][0]%mod)%mod;         ans=(ans+c[1]*res[0][1]%mod)%mod;         ans=(ans+81*res[0][2]%mod)%mod;         ans=(ans+27*res[0][3]%mod)%mod;         ans=(ans+9*res[0][4]%mod)%mod;         ans=(ans+3*res[0][5]%mod)%mod;         ans=(ans+res[0][6]%mod)%mod;         printf("%lld\n",ans);      }    }    return 0;}