hdu 1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46877    Accepted Submission(s): 10365


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<iostream>

using namespace std;

int main()

{

int a,b,n,c[50]; 

 while(cin>>a>>b>>n,a||b||n)

 { 

 c[0]=1;

c[1]=1;

for( int i = 2; i < 50; i++) 

 { 

 c[i]=(a*c[i-1]+b*c[i-2])%7;

}

if(n%49==0)

 cout<<c[48]<<endl; 

 }

else

 cout<<c[n%49-1]<<endl;

 }

 }

return 0;

}


#include <iostream>

using namespace std;

int main()

{

int A,B,fn,fn1,fn2,i,map[49];

int n; 

 while(cin>>A>>B>>n)

{

if(n==0&&B==n&&A==B) break;

map[0]=1;

map[1]=1;

for(i=2;i<=48;i++)

{ 

 map[i]=(A*map[i-1]+B*map[i-2])%7;

 }

if(n%49==0) 

cout<<map[48]<<endl; 

 else

{ 

 cout<<map[n%49-1]<<endl;

}

 }

return 0;

 }

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值