1005:Number Sequence

    技术2024-11-06  25

    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 代码: #include<cstdio> const int max=52; int f[max]; int  main() {  int a,b,n;  scanf("%d%d%d",&a,&b,&n);  a%=7;  b%=7;  while(a||b||n)  {   f[1]=1,f[2]=1;   for(int i=3;i<max;i++)   {    f[i]=(a*f[i-1]+b*f[i-2])%7;    if(f[i-1]==f[3]&&f[i]==f[4]&&i>4)     break;   }   int t=i-4;           if(n<4)    printf("%d/n",f[n]);          else    printf("%d/n",f[(n-4)%t+4]);  /* n=(n-2)%(i-4);   printf("%d/n",f[2+n]);*/   scanf("%d%d%d",&a,&b,&n);  }  return 0; }
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