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最新推荐文章于 2020-09-24 10:36:08 发布

翻译最新推荐文章于 2020-09-24 10:36:08 发布 · 369 阅读

文章标签：

#input #signal #each #output

Number Sequence

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

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

Sample Output

2
5

Author

CHEN, Shunbao

Source

ZJCPC2004

Recommend

JGShining

这道题看似简单，其实很纠结。因为 1 <= n <= 100,000,000，所以很有可能不是超时就是超内存。讨论区的大神们发帖子说每n%48一组结果是一个循环，算是找到了规律吧，然后就简单了。这题纯粹就是坑爹的。真不知道是怎么发现的。

#include<stdio.h>

#define M 49

int A;
int B;

int n;

int f[M];

int main(void)
{
    f[1]=1;
    f[2]=1;
    while(scanf("%d%d%d",&A,&B,&n),n){
        for(int i=3;i<49;i++)
            f[i]=(A*f[i-1]+B*f[i-2])%7;
        printf("%d\n",f[n%48]);
    }
    return 0;
}