HDU 1005:Number Sequence

HDU 1005:Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 141409    Accepted Submission(s): 34334

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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#includeint main(){ int a,b,n,z=1; int f[54]= {0,1,1}; while(scanf("%d%d%d", &a, &b, &n)!= EOF) { if(a==0&&b==0&&n==0)break; for(int i = 3; i < 54; ++i) { f[i]=(a*f[i-1]+b*f[i-2])%7; if(i>5) if(f[i-1]==f[3]&&f[i]==f[4]) { z=i-4; break; } } if(n>2)printf("%d\n",f[(n-3)%z+3]); else printf("1\n"); } return 0;}

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