HDU-1005 Number Sequence

最新推荐文章于 2020-07-12 20:18:12 发布

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本文介绍了一种解决特定数列问题的方法，该数列定义为f(1)=1,f(2)=1,f(n)=(A*f(n-1)+B*f(n-2))mod7，并通过示例展示了如何利用循环和取模运算高效计算f(n)。

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Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 191497 Accepted Submission(s): 47885

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 31 2 100 0 0

Sample Output

25

解答：两个数相加后对7取模，周期为7*7=49

代码：

#include<iostream>  
#include<queue>
#include<cstdio>  
#include<cstring>  
#include<algorithm> 
#include<vector> 
#include<cmath>  
#include<sstream>
#include<cstdlib>
#include<map>
const int N =1005;
using namespace std;
int main(){
	int	a,b,n,ans[50];
	ans[1] = ans[2] = 1;
	while(scanf("%d%d%d",&a,&b,&n),a!=0||b!=0||n!=0){
		for(int i =3;i<50;i++){
			ans[i] = (a*ans[i-1]+b*ans[i-2])%7;
		}
		cout<<ans[n%49]<<endl;
	}
	return 0;
}