HDOJ 1005

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 63888    Accepted Submission(s): 14701
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

我自己做的时候 忽略了n的大小，所以递归的时候栈溢出了.....

我的代码：

 1 #include <iostream>
 2 using namespace std;
 3 static int count;
 4 int f(int a,int b,int n);
 5 int mod(int n);
 6 int main()
 7 {
 8     int a,b,n;
 9     while(cin>>a>>b>>n)
10     {
11         if(a>=1 && a<=1000 && b>=1 && b<=1000)
12         {
13             count=0;
14             cout<<f(a,b,n)<<endl;
15         }
16         else
17             break;
18     }
19     return 0;
20 }
21 int f(int a,int b,int n)
22 {
23     if( n == 1 )
24         return 1;
25     else if( n == 2)
26         return 1;
27     else
28         return mod(a*f(a,b,n-1)+b*f(a,b,n-2));
29 }
30 int mod(int n)
31 {
32     while(n>7)
33     {
34         n=n-7;
35     }
36     return n;
37 }

网上大神的代码：

#include <iostream>
#include <vector>
using namespace std;
vector<int> ivec;
//布尔数组元素flag[i][j]如果为true,则说明前面已经出现了i 和 j两者的组合
bool flag[7][7];
void init(void)
{
    int i,j;
    for(i = 0;i < 7;++i)
        for(j = 0;j< 7;++j)
            flag[i][j] = false;
}
int main()
{
    int a,b,n;
    while(cin>>a>>b>>n)
    {
        if(a == 0&&b == 0&&n == 0)
            break;
        ivec.clear();
        init();
        ivec.push_back(1);
        ivec.push_back(1);
        flag[1][1] = true;
        int count = 1,f;
        while(1)
        {
            f = (a*ivec.at(count)%7 + b*ivec.at(count - 1)%7)%7;
            ivec.push_back(f);
            ++count;
            //如果flag变量为true，则说明前面已经出现了这两者的组合，出现重复，无需下一步计算，直接break退出即可
            if(flag[ivec.at(count)][ivec.at(count - 1)] == true)
                break;
            else
                flag[ivec.at(count)][ivec.at(count - 1)] = true;
        }
        //count中存放的是ivec中出现循环前的元素总个数,注意ivec中的下标是从0开始计数的
        count = count - 1;
        if(n < count)
            cout<<ivec.at(n-1)<<endl;
        else
        {
            int j;
            //for循环的目的是找出从那个地方开始重复，此处应该是从j处开始循环,注意j是从0下标开始计数的
            for(j = 0;;++j)
                if(ivec.at(count) == ivec.at(j) && ivec.at(count + 1) == ivec.at(j+1))
                    break;
            n = (n - j)%(count - j);
            if(n == 0)
                n = count - j;
            n += j;
            cout<<ivec.at(n-1)<<endl;
        }
    }
    return 0;
}