杭电acm 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 116886    Accepted Submission(s): 28442

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

 

Author

CHEN, Shunbao

 

 首先要知道mod是求余运算，一开始没考虑那么多 ，直接迭代，结果超时！ 仔细观察 对于这样的题目，通过第二个程序的打表代码不难发现肯定有输出规律，那么只需找出循环节即可，请参考大牛代码，我也学习了。

 

 

 

//一开始迭代 超时，通过打表可以看出循环#include<iostream>using namespace std;int main(){ int A,B,n; int a,a1,a2; while(cin>>A>>B>>n,A+B+n) {  a=0;a1=a2=1;  for(int i=0;i<n-2;i++)  {   a=(A*a1+B*a2);      a2=a1;   a1=a;  }  cout<<a<<endl; } return 0;}AC代码：#include<iostream>using namespace std;int f[100000005];int main(){    int a,b,n,i,j;    f[1]=1;f[2]=1;    while(cin>>a>>b>>n,a+b+n)    {        int s=0;//记录周期                for(i=3;i<=n;i++)        {            f[i]=(a*f[i-1]+b*f[i-2])%7;            for(j=2;j<i;j++)            if(f[i-1]==f[j-1]&&f[i]==f[j])//此题可以这样做的原因就是 2个确定后就可以决定后面的            {                s=i-j;                //cout<<j<<" "<<s<<" >>"<<i<<endl;                break;            }            if(s>0) break;//一开始s==0；如果找到循环节，即可跳出循环；        }        if(s>0){                 f[n]=f[(n-j)%s+2];//对于每个出入的n，转换为一个循环节内的数并复制给所求；                 //cout<<"f["<<n<<"]:="<<"f["<<(n-j)%s+j<<"] "<<endl;               }        cout<<f[n]<<endl;    }    return 0;}