杭大 1005 Number Sequence

Posted 2021-01-08 shenyuling

Number Sequence

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

原理：MOD7之后肯定有个周期，把他找出来，这题就完成了。具体如下

提示：周期是从第三个数开始，要不然会有下面这种情况

Sample input

7 7 10

则f[i]=1  1   0   0  0   0   0这种序列。很显然，周期是从第三个数开始循环的

 

#include<stdio.h>
int f[10000+5]={1,1,1};
int main()
{
    int a,b,n;
    while(scanf("%d%d%d",&a,&b,&n),a||b||n)
    {
        int i;
        if(n==2||n==1)
        {
            printf("1
");
            continue;
        }
        for(i=3;i<100;i++)//i<100表示，假定周期不会超过100左右，不过超了就改大一点呗
        {
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            if((i!=4&&f[i]==f[4]&&f[i-1]==f[3])||i-4>=n)
               break;
        }
        int T=i-2-2;//由于从第三个开始，所以周期T就有这样的表达式 
        printf("%d
",f[(n-3)%T+3]);//n-3=n-2-1;  n-2表示是周期的第n-2个数，再减一，是为了取模时数的范围在0 ——（T-1）
                                    //加三，是因为从第个开始
    }
    return 0;
}