HDU1005 Number Sequence（找规律，周期是变化的）

Posted 2020-12-03 yinbiao

传送门：

http://acm.hdu.edu.cn/showproblem.php?pid=1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 198316    Accepted Submission(s): 49744

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

 

题目意思：

输入a，b，n

根据a，b确定表达式，根据表达式确定f（n）

f（n）=(a*f(n-1)+b*f(n-2))%7

分析：

n太大了，不能递归，也不能打表

只能找规律

确定循环周期就好办了

记住循环周期随着输入是变化的

code：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b,n;
    int f[10000];
    f[1]=1;
    f[2]=1;
    while(~scanf("%d %d %d",&a,&b,&n))
    {
        if(a==0&&b==0&&n==0)
            break;
        int i;
        for(i=3;i<10000;i++)
        {
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if(f[i]==f[1]&&f[i-1]==1)//两个连续的1出现，表面循环到了一个周期的结束
                break;
        }
        n=n%(i-2);//周期为i-2
        f[0]=f[i-2];//周期结束的那个
        printf("%d
",f[n]);
    }
    return 0;
}