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hdu 1005 Number Sequence

时间:2015-02-07 17:11:19      阅读:157      评论:0      收藏:0      [点我收藏+]

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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
题目要求 就是f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) %7问你给你a,b,输出f(n);
因为n的范围不能直接递归,这是一个循环的解Ps题目测试数据有问题,真正意义上的ac代码不能过;
ac代码
 1 #include<stdio.h>
 2 #include<string.h>
 3 
 4 int a,b;
 5 int fun(int n)
 6 {
 7     if(n==1)
 8         return 1;
 9     else if(n==2)
10         return 1;
11     else
12         return (a*fun(n-1)+b*fun(n-2))%7;
13 }
14 
15 int main()
16 {
17     int n;
18     while(scanf("%d %d %d",&a,&b,&n)!=EOF&&(a!=0||b!=0||n!=0))
19     {
20 
21         printf("%d\n",fun(n%49));
22     }
23 }

实际情况的测试代码

#include<iostream>
#include<cstring>
using namespace std;

int x[100];
int main()
{
    int a,b,n,i,k;
    while(cin>>a>>b>>n,a||b||n)
    {
        memset(x,0,sizeof(x));
        x[1]=1,x[2]=1;
        for(i=3;i<60;i++)
        {
            x[i]=(a*x[i-1]+b*x[i-2])%7;
        }
        for(i=1;i<60;i++)
            cout<<i<< <<x[i]<<endl;
        for(i=3;i<60;i++)
            if(x[i]==1&&x[i+1]==1)
                break;
        k=i-1;
        if(n%k==0)
            cout<<x[k]<<endl;
        else
            cout<<x[n%k]<<endl;
        cout<<k<<endl;
    }
    return 0;
}

ac不带表你程序怎么样只代表这道题目不混过去了,只有写出真正解决实际问题的代码才是好代码

ps只是给你现实应该出现的数据

 

hdu 1005 Number Sequence

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原文地址:http://www.cnblogs.com/LQBZ/p/4278893.html

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