HDU-1005-Number Sequence

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Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 177757    Accepted Submission(s): 44122 Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.   Output For each test case, print the value of f(n) on a single line.   Sample Input 1 1 3 1 2 10 0 0 0   Sample Output 2 5

题解：这道题可以用矩阵来做，当然如果找规律也可以，用到了矩阵的快速幂，如果理解，对于很多题都有好处

通过观察可以看到这样一个关系，这样就能把这道题转化为矩阵快速幂了


。

#include<stdio.h>
#include<cstring>
using namespace std;
const int mod=7;
struct Matrix{ //用结构体比较好理解，如果对指针很熟练，用指针做形参也行
int f[2][2];
}mat;
Matrix Mul(Matrix a,Matrix b,int mod) //矩阵相乘
{
Matrix res;
memset(res.f,0,sizeof(res.f));
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
for(int k=0;k<2;k++)
res.f[i][j]=(res.f[i][j]+(a.f[i][k]%mod)*(b.f[k][j]%mod))%mod;
return res;
}
Matrix FastPower(Matrix mat,int k,int mod) //矩阵快速幂，与整数的快速幂思路一样
{
Matrix t;
memset(t.f,0,sizeof(t.f)); //清零
for(int i=0;i<2;i++)
t.f[i][i]=1;
while(k)
{
if(k&1) //如果k时奇数时，先让t乘一个，这样剩下的就是偶数个了
t=Mul(t,mat,mod); //t与任何矩阵相乘原矩阵都不变，还是mat
k/=2;
mat=Mul(mat,mat,mod);
}
return t;
}
int main()
{
int a,b,n;
while(~scanf("%d%d%d",&a,&b,&n))
{
if(a==0&&b==0&&n==0)
break;
if(n==1||n==2)
printf("1\n");
else
{ //赋初值
mat.f[0][0]=a;
mat.f[0][1]=b;
mat.f[1][0]=1;
mat.f[1][1]=0;
Matrix ans=FastPower(mat,n-2,mod);
printf("%d\n",(ans.f[0][1]+ans.f[0][0])%mod);
}
}
return 0;
}

Number Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/ b0e1 Others) Total Submission(s): 177757    Accepted Submission(s): 44122 Problem Description A number sequence is defined as follows: f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7. Given A, B, and n, you are to calculate the value of f(n).   Input The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.   Output For each test case, print the value of f(n) on a single line.   Sample Input 1 1 3 1 2 10 0 0 0   Sample Output 2 5