C Language problem solving

Update one every day OJ subject , Today I will introduce how to find the array sequence .

Preface

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).

The 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 need to calculate f（n） Value .

One 、 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.

The input consists of multiple test cases . Each test case contains 3 It's an integer A、B and n （1 <= A, B <= 1000, 1 <= n <= 100,000,000）. Three zeros indicate the end of input , This test case is not processed .

Two 、 Output

For each test case, print the value of f(n) on a single line.

For each test case , Print on one line f（n） Value .

Sample Input

1 1 3

1 2 10

0 0 0

 

Sample Output

2 5

Ideas ： The idea is simple , It's OK to find it directly in a cycle , But there's a problem , That's it n There's a lot of scope , This is the cycle of violence, then it is easy to timeout , And this question must not be as simple as it looks , First, we analyze a digital analog 7 There are seven possibilities 0,1,2,3,4,5,6; So the expression above can be f(n) = (A * f(n - 1) mod 7 + B * f(n - 2) mod 7) thus it can be seen ,f[n] Only 49 Medium condition , Then, according to a sample column, the pre 50 term , It can be found that the period is 49 Number of cycles , So just ask f[n%49] Just go .

Code

#include<stdio.h>
int main()
{
    int a,b,n;
    int f[100];
    f[0]=1;f[1]=1;
    while(1)
    {
        scanf("%d%d%d",&a,&b,&n);
        if (a < 1 || b > 1000 || n < 1 || n > 100000000)
            return -1;
        if(a==0&&b==0&&n==0)
           break;
        for(int i=2;i<49;i++)
            f[i]=(a*f[i-1]+b*f[i-2])%7;
        printf("%d\n",f[(n-1)%49]);
    }
    return 0;
}

That's what we're going to talk about today .