杭电acm_1005Number Sequence

缺乏、安全感 2021-12-13 03:45 256阅读 0赞

[http://acm.hdu.edu.cn/showproblem.php?pid=1005][http_acm.hdu.edu.cn_showproblem.php_pid_1005]

Number Sequence  
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 37400    Accepted Submission(s): 7879

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

#include <stdio.h>
    int main()
    {
        int a,b;
        long n;
        int f[201];
        int i;
        while(scanf("%d%d%ld",&a,&b,&n)!=EOF)
        {
            if(n==0&&a==0&&b==0)
                break;
            f[1]=1;
            f[2]=1;
            if (n>=3)
            {
                for(i=3;i<200;i++)
                {
                    f[i]=(a*f[i-1]+b*f[i-2])%7;
                    if ((f[i-1]==1)&&(f[i]==1))
                    {
                        break;
                    }
                }
                i=i-2;
                n%=i;
                if (n==0)
                {
                    n=i;
                }
                printf("%d\n",f[n]);
            }
            else
            {
                printf("1\n");
            }
        }
        return 0;
    }

转载于:https://www.cnblogs.com/pcoda/archive/2011/04/26/2104599.html

[http_acm.hdu.edu.cn_showproblem.php_pid_1005]: http://acm.hdu.edu.cn/showproblem.php?pid=1005