杭电ACM 1005 Number Sequence

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

 

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给出的计算表达式里面有f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.由于这个%符号的存在，所以f（n）的取值只有0-6七种，所以f(n-1)和f(n-2)的取值最多有49种，所以f(n)是个周期数列，并且周期不大于49。由上面思路可以写如下代码：

/************************************************************************/
/* 1005 Number Sequence                                                */
/*由f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.    */
/*主要是mod7可以得到f（n）的周期不会超过49                             */
/************************************************************************/
#include <iostream>

using namespace std;
int f[60];
int T;
int main()
{
int a, b ,n;
f[1] = 1;
f[2] = 1;
while (cin>>a>>b>>n && a && b && n )
{
if (n==1 || n==2)
{
cout << f
<< endl;
}
else
{
for (int i = 3; i < 60; i++)
{
f[i] = (a*f[i - 1] + b*f[i - 2]) % 7;
if (f[i] == 1 && f[i - 1] == 1)
{
T = i ;
break;
}
}
if (n>T)
{
n = n % (T-2);
if (n==0)
{
n = T-2;
}
}
cout << f
<< endl;
}
}
return 0;
}

但是发现一个问题：将T=T-2放在if(n>T)之前会发生无法ac的情况，跪求大神和各位兄弟姐妹指导。代码如下：

/************************************************************************/
/* 1005 Number Sequence                                                */
/*由f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.    */
/*主要是mod7可以得到f（n）的周期不会超过49                             */
/************************************************************************/
#include <iostream>

using namespace std;
int f[60];
int T;
int main()
{
int a, b ,n;
f[1] = 1;
f[2] = 1;
while (cin>>a>>b>>n && a && b && n )
{
if (n==1 || n==2)
{
cout << f
<< endl;
}
else
{
for (int i = 3; i < 60; i++)
{
f[i] = (a*f[i - 1] + b*f[i - 2]) % 7;
if (f[i] == 1 && f[i - 1] == 1)
{
T = i ;
break;
}
}
T = T - 2;
if (n>T)
{
n = n % T;
if (n==0)
{
n = T;
}
}
cout << f
<< endl;
}
}
return 0;
}