hdu 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60820    Accepted Submission(s): 13862

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

Author

CHEN, Shunbao

Source

ZJCPC2004

Recommend

JGShining

 

 

 

这有个我十分的不解，这个程序整整RE了很多次。可是和我AC的程序只是将len改成i-2

不信你可以试试。

解题思路是找出循环节；

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 100000
using namespace std;
int f[N];
int main()
{
  f[1]=f[2]=1;
  int a,b,n;
  int len;
  while(cin>>a>>b>>n,a||b||n)
  {
    int i;
    for(i=3;i<10000;i++)
    {
      f[i]=(a*f[i-1]+b*f[i-2])%7;
      if(f[i]==1&&f[i-1]==1)
        {
          len=i-2;
          break;
        }
    }
    n=n%len;
    f[0]=f[len];
    cout<<f[n]<<endl;
  }
  return 0;
}

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 100000
using namespace std;
int f[N];
int main()
{
  f[1]=f[2]=1;
  int a,b,n;
  int len;
  while(cin>>a>>b>>n,a||b||n)
  {
    int i;
    for(i=3;i<10000;i++)
    {
      f[i]=(a*f[i-1]+b*f[i-2])%7;
      if(f[i]==1&&f[i-1]==1)
        {
          break;
        }
    }
    n=n%(i-2);
    f[0]=f[i-2];
    cout<<f[n]<<endl;
  }
  return 0;
}