HDU 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 178891    Accepted Submission(s): 44454

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

 題解：我看這道題的時候，第一眼就看見n的範圍很大，所以，常規的遞歸肯定會超時，所以這是一道規律題……因爲結果跟A B有關，所以x[i]的值肯定跟x[i-1]和x[i-2]有關，我猜測1 ，1 一定是這段循環體的開頭，所以我就用for循環找，直到從新找到1,1這個循環體的開頭，進而求出循環體的長度，然後n對長度取餘，進而求解，但是提交之後wrong了……

後來，我寫了一段代碼，模擬了當A，B任意取值時，x[i]的記過。發現了一段特殊的循環體“0”，當Ａ，Ｂ的值都是７的倍數時，x[i]的值從第三項起都是0，我把這個特殊的循環體給特意寫出來就直接過了……

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<string>
#include<math.h>
#include<map>
#include<queue>
#include<stack>
#define INF 0x3f3f3f3f
#define ll long long
#define f(i,a,b) for(int i=a;i<b;i++)
#define sf(a)  scanf("%d",&a)
#define sff(a,b)  scanf("%d%d",&a,&b)
#define sfff(a,b,c)  scanf("%d%d%d",&a,&b,&c)
#define pf(a) printf("%d\n",a)
#define p printf("\n")
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
#define mod 1000000
int x[10000];
int main()
{
    int a,b,n,c=1;
    while(~sfff(a,b,n)&&(a,b,n))
    {
        if(a%7==0&&b%7==0)
        {
            if(n>2)pf(0);
            if(n<=2)pf(1);
            continue;
        }
        x[1]=1,x[2]=1;
        f(i,3,10000)
        {
            x[i]=(a*x[i-1]+x[i-2]*b)%7;
            if(x[i]==1)
            {
                if(x[i-1]==1)
                {
                    c=i-2;
                    break;
                }
            }
        }
        int dd=n%c;
        if(dd)pf(x[dd]);
        else pf(x[c]);
    }
}