hdu 1005 Number Sequence

不念不忘少年蓝@ 2022-06-16 01:14 147阅读 0赞

# Number Sequence #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 170784 Accepted Submission(s): 42133**

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
      
       
      
       
        
       题目解析：这个题的意思就是要求
       
        f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.的值。
      
       
      
       
       
        
    
      
       
      
       
       
        题目解析：直接对这个方程用函数或递归直接写出都会导致内存超出，要考虑一下，经过思考后发现，这个题会出现循环，48是他的循环周期，这样这个题就变得很简单了。
      
       
      
       
       
        
    
      
       
      
       
       
        代码：
      
       
      
       
       
        
       
        #include<cstdio>
    int ans[49];
    int main()
    {
        int A,B,n,i;
        while(~scanf("%d%d%d",&A,&B,&n))
        {
            if(A==0&&B==0&&n==0)
                printf("");
            else
            {
                ans[0]=0;
                ans[1]=1;
                ans[2]=1;
                for(i=3; i<=48; i++)
                    ans[i]=(A*ans[i-1]+B*ans[i-2])%7;
    
                printf("%d\n",ans[n%48]);
            }
        }
        return 0;
    }