HDU 1005 Number Sequence
2015-03-15 13:48
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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
思路:这道题刚开始用普通的方法,数组开了100000000会内存溢出,后来用两个变量循环使用解决这一问题,又超时了。显然,用普通的方法是行不通的。那么,我们在数学上遇到这种问题会想到找规律。对7取模,那么f(i)的值定在0-6之间,f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7可以写成f(n) = (A * f(n - 1)mod 7 + B * f(n - 2)mod 7) mod 7,也就是说,无论a,b取多少,都可以转化成0-6之间的范围,得到的结果是一样 的。所以a,b的组合共有7*7 = 49种。也就是说,f(i)是一个周期函数,且周期定在49以内。我们只要求出0-49的值,求出周期,就可以了。那么,周期如何求呢?注意到该函数,当前值与前两个值是相关的,若出现周期,定会从1,1,开始,而出现了1,1后面的值都能确定了。结论:只要判断相邻两个数是1,1即可。最后还要注意的是,我们求出了cycle(周期),但是n% cycle会等于0,还需要我们把发f(0) 求出。
代码:
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
思路:这道题刚开始用普通的方法,数组开了100000000会内存溢出,后来用两个变量循环使用解决这一问题,又超时了。显然,用普通的方法是行不通的。那么,我们在数学上遇到这种问题会想到找规律。对7取模,那么f(i)的值定在0-6之间,f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7可以写成f(n) = (A * f(n - 1)mod 7 + B * f(n - 2)mod 7) mod 7,也就是说,无论a,b取多少,都可以转化成0-6之间的范围,得到的结果是一样 的。所以a,b的组合共有7*7 = 49种。也就是说,f(i)是一个周期函数,且周期定在49以内。我们只要求出0-49的值,求出周期,就可以了。那么,周期如何求呢?注意到该函数,当前值与前两个值是相关的,若出现周期,定会从1,1,开始,而出现了1,1后面的值都能确定了。结论:只要判断相邻两个数是1,1即可。最后还要注意的是,我们求出了cycle(周期),但是n% cycle会等于0,还需要我们把发f(0) 求出。
代码:
#include <iostream> #include<cstdio> #include<cstring> using namespace std; int lis[59]; int main() { int a,b,n,i,cycle; while (scanf("%d %d %d",&a,&b,&n) && a && b && n) { memset(lis,0,sizeof(lis)); lis[1] = 1; lis[2] = 1; if (n <= 2) { cout << 1 << endl; } else { for (i = 3; i <= 49; i++) { lis[i] = (a * lis[i-1] + b * lis[i-2]) % 7; if(lis[i] == 1 && lis[i-1] == 1) { break; } } if (i >= n) { cout << lis[n] << endl; } else { cycle = i-2; lis[0] = lis[cycle]; cout << lis[n % cycle]<< endl; } } } return 0; }
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