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 31 2 100 0 0

 

Sample Output

25

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matk/=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/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 31 2 100 0 0

 

Sample Output

25