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hdu1005

发布时间:2022-07-27 13:56:07

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 72091    Accepted Submission(s): 16791 hdu1005
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                
  1. #include<iostream>
  2. using namespace std;
  3. int f[10000];
  4. int main()
  5. {
  6.  int a,b,n;
  7.  while(scanf("%d%d%d",&a,&b,&n)!=EOF)
  8.  {
  9.   int i;
  10.   if(a==b&&b==0&&n==0)
  11.    break;
  12.   f[1]=1;
  13.   f[2]=1;
  14.   for(i=3;i<48;i++)
  15.   {
  16.    f[i] = (a*f[i-1] +b*f[i-2])%7;
  17.   }
  18.   n=n%48;
  19.     printf("%d " ,f[n]);
  20.  }
  21.  return 0;
  22. }
 

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