hdoj1005_number sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 105414    Accepted Submission(s): 25611

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

 

 存在循環

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

struct Data
{
    bool f;
    int wei;
}flag[100];
int res[110];

int main()
{
    for (int i = 0; i < 100; i++)
    {
        flag[i].f = false;
        flag[i].wei = 0;
    }
    int a, b, n;
    while (scanf("%d %d %d", &a, &b, &n) != EOF && a && b && n)
    {
        for (int i = 0; i < 100; i++)
        {
            flag[i].f = false;
            flag[i].wei = 0;
        }
        memset(res, 0, sizeof(res));
        res[1] = 1;
        res[2] = 1;
        int len = 1, num = 3;
        bool f = false;
        flag[11].f = true;
        flag[11].wei = 1;
        int wei;
        while (!f)
        {
            res[num] = (a * res[num - 1] + b * res[num - 2]) % 7;
            int temp = res[num] * 10 + res[num - 1];
            if(flag[temp].f)
            {
                f = true;
                len = num - 2 - flag[temp].wei + 1;
                wei = flag[temp].wei;
            }
            else 
            {
                flag[temp].f = true;
                flag[temp].wei = num - 1;
                num++;
            }
        }
        if (n <= wei - 1)
            printf("%d\n", res[n]);
        else 
        {
            int r = (n - wei + 1) % len;
            if (r == 0)
                printf("%d\n", res[wei + len - 1]);
            else 
                printf("%d\n", res[wei + r - 1]);
        }

    }
    return 0;
}