HDOJ 1005 Number Sequence
2016-02-14 16:56
369 查看
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
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<iostream> #include<stdio.h> using namespace std; int f[100000005]; int main() { int a,b,n,i,j; f[1]=1;f[2]=1; while(scanf("%d%d%d",&a,&b,&n)) { int s=0;//记录周期 if(a==0&&b==0&&n==0) break; for(i=3;i<=n;i++) { f[i]=(a*f[i-1]+b*f[i-2])%7; for(j=2;j<i;j++) if(f[i-1]==f[j-1]&&f[i]==f[j])//此题可以这样做的原因就是 2个确定后就可以决定后面的 { s=i-j; //cout<<j<<" "<<s<<" >>"<<i<<endl; break; } if(s>0) break; } if(s>0){ f =f[(n-j)%s+j]; //cout<<"f["<<n<<"]:="<<"f["<<(n-j)%s+j<<"] "<<endl; } cout<<f <<endl; } return 0; }
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(sc.hasNext()){ int[] a = new int[54]; int A = sc.nextInt(); int B = sc.nextInt(); int n = sc.nextInt(); if(A==0&&B==0&n==0){ return ; } a[1]=1; a[2]=1; int k=0; int i; for(i=3;i<54;i++){ a[i] = (A*a[i-1]+B*a[i-2])%7; if(i>5&&a[i]==a[4]&&a[i-1]==a[3]){ k=i-4; break; } } //System.out.println(k); if(n>2){ System.out.println(a[(n-3)%k+3]); }else{ System.out.println("1"); } } } }
相关文章推荐
- QuickText | 热字串自动替换
- Codeforces 626A Robot Sequence 【水题】
- poj 2926 Requirements 【5维点集最远曼哈顿距离】
- Could not build module 'UIKit'问题
- PHP中include()与require()的区别说明
- hdu1530 Maximum Clique
- Arduino代码机制-Arduino.h
- Android开发中无处不在的设计模式——Builder模式
- POJ 2524 Ubiquitous Religions
- Longest Increasing Subsequence
- ruiaijun人工智能理论纪录片:《反射算法与离散智能编程》《自编程自主学习算法》
- poj2926 Requirements
- API Guides(二)——Activity To AIDL
- iOS7 iOS8 UITableviewCell处于编辑状态,dismiss或者back崩溃
- WeUI logo专为微信设计的 UI 库 WeUI
- 在一般处理文件中访问Session需要添加IRequiresSessionState
- BlockingQueue简介
- 编译ionic项目的时候提示: Please install the Android build tools version 19.1.0 or higher.
- squirrel sql client 连接phoenix
- EasyUi 中datagrid 实现查询方法