Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 125495    Accepted Submission(s):
30510

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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 1 #include <stdio.h>
 2 int num[55] ;   //数组开的要大于 i 最大值 ；
 3 int main()
 4 {
 5     int a,b,c,i,temp ;
 6     num[1] = 1 ; num[2] = 1 ;
 7     while(~scanf("%d %d %d",&a, &b, &c) , (a || b || c) )
 8     {
 9
10         for(i=3; i<=49; i++)
11         {
12             num[i] = (a*num[i-1] + b*num[i-2]) %7 ;
13             if(num[i] == num[i-1] && num[i] == 1 )
14             break ;
15         }
16         temp = i-2 ;
17         c = c%temp ;
18         if(c == 0)
19             printf("%d\n",num[temp]) ;  //  一直理解错，取余后得到0为循环最后一个，不是第一个；
20         else
21             printf("%d\n", num[c]) ;
22     }
23     return 0 ;
24 }