杭电1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 62216 Accepted Submission(s): 14226



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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JGShining

#include <iostream>
#include <vector>
using namespace std;

void fn(int A, int B, int n);
void fn(int A, int B, int n)
{

int cirlce[50];
cirlce[1] = 1;
cirlce[2] = 1;
if (n == 1 || n == 2)
{
cout << 1 << endl;
return;
}
int i;
for ( i = 3;i < 50 ; ++ i) //找到循环节
{
cirlce[i] = (A*cirlce[i -1] + B*cirlce[i -2])%7;
if (cirlce[i -1] == 1 && cirlce[ i ] ==1)
break;
}
if (n%(i -2) == 0)
{
cout << cirlce[i -2]<< endl;
}
else
{
cout << cirlce[n%(i -2)] << endl;
}
}

int main(int argc, char **argv)
{
int A, B, n;
while (cin>>A>>B>>n) {
if (A == 0 && B == 0 && n == 0)
break;
fn(A, B, n);

}
return 0;
}