首页 > 代码库 > Hdu 1005 Number Sequence
Hdu 1005 Number Sequence
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 173150 Accepted Submission(s): 42786
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> # include <string>int fun(int a, int b, int n){ if (n==1) return 1; else if(n==2) return 1; else return (a*fun(a, b, n-1)+b*fun(a, b, n-2))%7; } int main(){ int a, b, n; while(scanf("%d%d%d",&a,&b,&n)){ if(a == 0 && b == 0 && n == 0) break; else printf("%d\n",fun(a, b, n%49)); } system("pause"); return 0; }
Hdu 1005 Number Sequence
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。