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


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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.
 

[align=left]Sample Input[/align]

2
3 1 2
4 1 10

 

[align=left]Sample Output[/align]

85
369
HintIn 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 7
using namespace std;
typedef long long ll;
const ll mod=2147493647;
ll a

,temp

,res

,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

,ll b

)
{
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

,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;
}