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  • hdu1005 Number Sequence(数论)

     

    Number Sequence

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 86102 Accepted Submission(s): 20423

    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

    当时别人问我这题,我最开始一看感觉是矩阵快速幂,回来后想想%7,因为7太小了,可能有规律,周期最大为48,突然想起来原来以前做过这个题

    #include <stdio.h>
    int main()
    {
    	int i,j,A,B,f1,f2,f3,n;
    	while(scanf("%d%d%d",&A,&B,&n)!=EOF&&(A||B||n))
    	{
    		f1=1;
    		f2=1;
    		A=A%7;
    		B=B%7;
    		n=(n-3)%48+2;
    		for(i=2;i<=n;i++)
    		{
    			f3=(A*f2+B*f1)%7;
    			f1=f2;
    			f2=f3;
    		}
    		printf("%d
    ",f2);
    	}
    	return 0;
    }


     
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  • 原文地址:https://www.cnblogs.com/james1207/p/3366081.html
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