HDU 1005 Number Sequence (数学规律)

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 104190    Accepted Submission(s): 25232

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  

 

 

Author

CHEN, Shunbao

 

 

Source

ZJCPC2004

 

 

Recommend

JGShining

 

 

看了大神的代码才知道水题也是要很严谨去对待的啊。

题目主要是要找循环节。

一开始找规律于是又了下面的代码

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<stdlib.h>

 4 #include<algorithm>

 5 using namespace std;

 6 int a[100];

 7 int main()

 8 {

 9     //freopen("in.txt","r",stdin);

10     int n,A,B;

11     while(scanf("%d %d %d",&A,&B,&n)&&(n||A||B))

12     {

13         a[1]=1;

14         a[2]=1;

15         for(int i=3;i<=49;i++)

16             a[i]=(A*a[i-1]+B*a[i-2])%7;

17         printf("%d\n",a[n%49]);

18     }

19     return 0;

20 }

View Code

后来发现如果A=7，B=7，n>2的话，输入都应该会是0，但是实际输出为1。虽然数据有点水，所以还是要严谨对待。

做法还是找循环节，只是说多了很多细节，要好好体会。

 1 #include<cstdio>

 2 #include<cmath>

 3 #include<cstring>

 4 #include<stdlib.h>

 5 #include<algorithm>

 6 using namespace std;

 7 const int MAXN=200+10;

 8 int a[MAXN];

 9 int main()

10 {

11     //freopen("in.txt","r",stdin);

12     int A,B,n,i;

13     a[1]=a[2]=1;

14     while(scanf("%d %d %d",&A,&B,&n)!=EOF)

15     {

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

17         int flag=0;

18         for(i=3;i<=200;i++)

19         {

20             a[i]=(A*a[i-1]+B*a[i-2])%7;

21             if(a[i]==1&&a[i-1]==1) 

22                 break;

23             if(a[i]==0&&a[i-1]==0){

24                 flag=1;break;

25             }   

26         }

27         if(flag)

28         {

29             printf("0\n");

30             continue;

31         }

32         if(i>n)

33         {

34             printf("%d\n",a[n]);

35             continue;

36         }

37         i=i-2;

38         n=n%i;

39         if(n==0)n=i;

40         printf("%d\n",a[n]);

41     }

42     return 0;

43 }

View Code