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  • HDU_1005 Number Sequence

    Number Sequence

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 40967    Accepted Submission(s): 8785


    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
     

    找到规律,打表就可以了

    View Code
    #include <stdio.h>
    int f[1009];
    int main()
    {
    int a, b, i, n, c;
    f[
    1] = 1;
    f[
    2] = 1;
    while(scanf("%d%d%d",&a,&b,&n) && (a || b || n))
    {
    for(i = 3; i < 50;i ++)
    {
    f[i]
    = (a*f[i-1] + b*f[i-2]) % 7;
    if(f[i] == 1 && f[i-1] == 1)
    break;
    }
    c
    =i-2;
    n
    =n%c;
    if(n == 0)
    printf(
    "%d\n",f[c]);
    else
    printf(
    "%d\n",f[n]);
    }
    return 0;
    }
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  • 原文地址:https://www.cnblogs.com/vongang/p/2115046.html
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