Number Sequence

最新推荐文章于 2024-08-07 14:16:33 发布

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最新推荐文章于 2024-08-07 14:16:33 发布

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本文介绍了一种通过递推公式计算特定数列第n项值的算法，该数列涉及模运算并限定于7的取余结果。文章提供了一个完整的C语言实现案例，通过优化循环避免重复计算，有效地解决了大规模数值下的计算问题。

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1960 Accepted Submission(s): 734

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

ZJCPC2004

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#include<stdio.h>
#include<math.h>
#include<string.h>
int f[1000];
int main()
{
    int a, b;
    int n, i;
    while(1)
    {
        memset(f, 0, sizeof(f));
        scanf("%d%d%d", &a, &b, &n);
        if(a==0 && b==0 && n == 0) break;
        f[0] = f[1] = 1;
        if(n == 1 || n ==2 )
        {
            printf("1\n");
            continue;
        }
        for(i = 2; i<1000; i++)
        {
            f[i] = (a * f[i-1] + b * f[i-2])% 7;
            if(f[i-1] == 1 && f[i-2] == 1 && i != 2  )  break;
        }
        printf("%d\n",f[(n-1)%(i-2)]);
    }
    return 0;
}