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  • hdoj--1005--Number Sequence(规律题)

    Number Sequence

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 137047    Accepted Submission(s): 33211



    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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    规律题,48一循环

    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    using namespace std;
    int a[48];
    int main()
    {
    	int A,B,n;
    	while(scanf("%d%d%d",&A,&B,&n)!=EOF)
    	{
    		if(A==0&&B==0&&n==0) break;
    		memset(a,0,sizeof(a));
    		a[2]=a[1]=a[3]=1;
    		int t;
    		for(int i=3;i<=48;i++)
    		a[i]=(A*a[i-1]%7+B*a[i-2]%7)%7;
    		printf("%d
    ",a[n%48]);
    	}
    	return 0;
    }


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  • 原文地址:https://www.cnblogs.com/playboy307/p/5273657.html
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