Links: http://poj.org/problem?id=2440
Dna
Time limit:1000ms Memory limit:65536ktotal submissions:3498 accepted:1375
Description
A kind of virus has attacked the X planet, and many lives is infected. After weeks of study, the CHO (Creature Healthy Organization) of X Planet finally finds out that this kind of virus have TW O kind of very simple DNA, and can is represented by 101 and 111. Unfortunately, the lives on the planet also has DNA formed by 0s and 1s. If a creature ' s DNA contains the virus ' DNA, it'll be affected; Otherwise it would not. Given an integer l, it's clear that there'll be 2 ^ L different lives, of which the length of DNA is L. Your job is-find out in the 2 ^ L lives how many won ' t be affected?
Input
The input contains several test cases. For each test case it contains a positive integer l (1 <= L <= 10 ^ 8). The end of input is indicated by End-of-file.
Output
For each test case, the output K mod 2005, here K was the number of lives that would not be affected.
Sample Input4
Sample Output9
Source
POJ monthly,static
The main idea-a virus that invades a planet, which has two kinds of DNA, 111 and 101, which has an L-bit gene and is made up of 0 and 1. If a biological gene contains the same substring as the virus, it will be affected. Q: Given an L, how many possible combinations of genes will not be affected, resulting in 2005 modulo.
Thought--l value is larger than the largest, the maximum is 10^8, but the result to 2005 modulo, the explanation must exist the Circulation section, and must have the recursion relation. Note: F (0) =1,f (1) =2,f (2) =4,f (3) =6,f (4) =9,f (5) =15,f (6) =25,f (7) = 40, thus obtaining f (n) =f (n-1) +f (n-3) +f (n-4), n>3. Then the recursive relationship can be used to find the loop section to solve.
Complexity analysis--time complexity: O (n), spatial complexity: O (n)
Attach the AC code:
#include <iostream> #include <cstdio> #include <string> #include <cmath> #include <iomanip > #include <ctime> #include <climits> #include <cstdlib> #include <cstring> #include < algorithm>using namespace std;typedef unsigned int ui;typedef long long ll;typedef unsigned long long ull;typedef long Double Ld;const double PI = 3.14159265;const double E = 2.71828182846;const int mod = 2005;const int look = 1000;int fib[l Ook] = {1, 2, 4, 6, 9, 15};int Main () {Ios::sync_with_stdio (false); int loop;for (int i=6; i<=look; i++) {Fib[i] = (fib[i- 1]+FIB[I-3]+FIB[I-4])%2005;if (fib[i] = = 2 && fib[i-1] = = 1) {loop = I-1;break;}} int L;while (cin >> L) {cout << fib[l%loop] << Endl;} return 0;}
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POJ 2440 DNA
