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  • 杭电1005--Number Sequence

    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
     
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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
    ",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
    ", num[c]) ;
    22     }
    23     return 0 ;
    24 } 
     
     
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  • 原文地址:https://www.cnblogs.com/soTired/p/4634038.html
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