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HDU 1005 Number Sequence (数学规律)

2024-07-18 10:54:40   232人阅读

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 31 2 100 0 0
 

 

Sample Output
25 
 
 
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 }
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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 }
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