hdu1005Number Sequence(循环节)

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 154006    Accepted Submission(s): 37586

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

 

Source

第一种方法：找出循环节s，

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int f[100000001];int main(){    f[1]=1;    f[2]=1;    int i,n,j;    int a,b;    while(scanf("%d%d%d",&a,&b,&n)!=EOF&&(a||b||n))    {        int s=0;        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])                {                    s=i-j;                    break;                }            if(s>0)break;        }        if(s>0)        {            f[n]=f[(n-j)%s+j];        }        cout<<f[n]<<endl;;    }    return 0;}

第二种，由于最大mod为7，由鸽巢原理知最大状态数不超过7^7

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <set>using namespace std;int main(){    int f[1001],n;    int a,b,i;    while(scanf("%d%d%d",&a,&b,&n)&&a+b+n)    {        f[1]=1;        f[2]=1;        for(i=3;i<=49;i++)        {            f[i]=(a*f[i-1]+b*f[i-2])%7;        }        printf("%d\n",f[n%48]);    }    return 0;}