HDU 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 129265    Accepted Submission(s): 31474

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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既然模7了结果只有0到6，七种可能，所以猜想是有循环节的，但是不同的a,b循环点不一样。所以自己找一下循环点， 7 7数据特殊结果全是0，单独处理。

#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 105
#define INF 0x3f3f3f3f

int a,b,n,mod;
__int64 f[N];

int main()
{
    f[1]=1;f[2]=1;
    while(~scanf("%d%d%d",&a,&b,&n)&&(a+b+n))
    {
        if(a==7&&b==7)
        {
            printf("0
");
            continue;
        }
        for(int i=3;i<=100;i++)
        {
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if(i>6 && f[i]==f[3] && f[i-1]==1 &&f[i-2]==1)
            {
                mod=i-3;break;
            }
        }
        n=n%mod;
        if(n==0)n=mod;
        cout<<f[n]<<endl;
    }
    return 0;
}