HDU 1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 94119    Accepted Submission(s): 22487

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

ZJCPC2004

 

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思想：

由于n值太大，所以需要削减循环次数。注意到%7，所以f(n)的值只可能为0 1 2 3 4 5 6其中的某一个，由于f1=1, f2=1从而可以得到f3,f4,f5等等。

当其中后面的某连续两项同为1时，即：f(k-1) = 1, f(k) = 1  这时候又回到了f1, f2。

/*Title：HDU 1005 Author：Dojking*/#include <stdio.h>int find(int A, int B)   //找出循环区间 {int n = 0, f1 = 1, f2 = 1, f3 = 1;while (!(1 == f1 && 1 == f2) || !n){f3 = (A*f2+B*f1)%7;f1 = f2;f2 = f3;//printf("n =  %d,  f1 = %d,  f2 = %d\n",n, f1, f2);++n;}return n;} int main(){int i, mod, n, A, B, f1, f2, f3;while (scanf("%d%d%d", &A, &B, &n)==3 && (A && B && n)){mod = find(A, B);n %= mod;//printf("mod = %d,  n = %d\n",mod, n);f3 = 1;f1 = 1;f2 = 1;for (i = 3; i <= n; ++i)   //求出f(n)的值，即f3 {f3 = (A*f2+B*f1)%7;f1 = f2;f2 = f3;}printf("%d\n",f3);}return 0;}