【hdoj1005】Number Sequence

最新推荐文章于 2023-05-04 21:56:48 发布

原创最新推荐文章于 2023-05-04 21:56:48 发布 · 280 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#规律

HDOJ 同时被 2 个专栏收录

111 篇文章

订阅专栏

易超时

3 篇文章

订阅专栏

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 158167 Accepted Submission(s): 38744

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

Sample Output

2
5

Author

CHEN, Shunbao

Source

ZJCPC2004

Recommend

JGShining

规律是48个一个循环，直接打表前48个，再取余一下就好了。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mod 7
using namespace std;
typedef long long LL; 
const int N = 100;
int ans[N];
void init(int A,int B){
	ans[1]=1,ans[2]=1;
	for(int i=3;i<=48;i++){
		ans[i]=(A*ans[i-1]%mod+B*ans[i-2]%7)%7;
	}
}
int main(){
	LL A,B,n;
	while(scanf("%lld%lld%lld",&A,&B,&n),A||B||n){
		init(A,B);
		printf("%d\n",ans[n%48]);
	}
	return 0;
}