hduOJ1005 Number Sequence //算法

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 202453    Accepted Submission(s): 50972

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

    我开始想用递归直接就求出来了（天真），但是提交后才知道Runtime Error(INTEGER_DIVIDE_BY_ZERO)；

n太大可能递归次数多就溢出了。每个线程都建立一个默认1M堆栈。这个堆栈是供函数调用时使用，线程内函数里的各种静态变量都是从这个默认堆栈里分配的.

RE代码：

 1 #include<iostream>
 2 using namespace std;
 3 int f(int a,int b,int n) {    
 4     int m;
 5     if(n==1||n==2)
 6         return 1;
 7     else {
 8         m=(a*f(a,b,n-1)+b*f(a,b,n-2))%7;　　　　//会溢出
 9         return m;
10     }
11 }
12 int main() {
13     while(1) {
14         int A,B,n;
15         cin>>A>>B>>n;
16         if(A==0&&B==0&&n==0||A<1||B<1||n<1)
17             break;
18         else {
19             cout<<f(A,B,n)<<endl;
20         }
21     }
22     return 0;
23 }

还是看了别人的才发现mod7是有规律的，周期是49，所以：

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int f[105];
 5 int main()
 6 {
 7     int a,b,n;
 8     while(1)
 9     {
10         cin>>a>>b>>n;
11         if(a==0&&b==0&&n==0)
12         break;
13         f[1]=1,f[2]=1;
14         for(int i=3;i<=49;i++)
15         {
16             f[i]=(a*f[i-1]+b*f[i-2])%7;
17         }
18         int num=n%49;
19         cout<<f[num]<<endl;
20     }
21     return 0;
22 }