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HDU1005

Number Sequence  HDU-1005

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 162730    Accepted Submission(s): 40016

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
 
 
发这道题主要是祭奠一下我第一次遇到的超内存MLE。。。。。第一想法这么简单,暴力递归。。然后MLE狠狠打脸。。

后来再认真读题,发现还是有规律的。

如果f(n - 1)和f(n)都为1的话,就是一个循环。

#include <iostream>
#include <cstdio>
using namespace std; int f[51]; int main() { int a, b; long long n; while (scanf("%d %d %lld", &a, &b, &n) == 3) { if (a == 0 && b == 0 && n == 0) break; f[1] = f[2] = 1; int i; for (i = 3; i < 51; i++) { f[i] = (a * f[i - 1] + b * f[i - 2]) % 7; if (f[i] == 1 && f[i - 1] == 1) //找到循环因子 i { break; } } n = n % (i - 2); if (n == 0) //刚好经过一个循环 printf("%d\n", f[i - 2]); else printf("%d\n", f[n]); } return 0; }

这题真的是好烦。。。。。。。

 

HDU1005