HDU 1005 Number Sequence 找规律

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

 

 

/*
    一开始没看到n<=10000000
    结果疯狂TLE
    
    最后看了discuss 才知道要找周期 
*/
#include<cstdio>
#include<iostream>
#define MAXN 100000010

using namespace std;

int a,b,n;

int f[1001];

int main() {
    while(~scanf("%d %d %d",&a,&b,&n)&&a&&b&&n) {
        int pos;
        f[0]=0;f[1]=1;f[2]=1;
        for(int i=3;i<=100;i++) {
            f[i]=(f[i-1]*a+f[i-2]*b)%7;
            if(f[i]==f[2]&&f[i-1]==f[1]) {  //找到周期 
                pos=i;
                break;
            }
        }
//        printf("%d\n",pos-2);
        n=n%(pos-2);  //周期为pos-2 
        if(n!=0) printf("%d\n",f[n]);
        else printf("%d\n",f[pos-2]);
    }
    return 0;
}

 

 

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