HDU 3485 Count 101

Problem Description

You know YaoYao is fond of his chains. He has a lot of chains and each chain has n diamonds on it. There are two kinds of diamonds, labeled 0 and 1. We can write down the label of diamonds on a chain. So each chain can be written as a sequence consisting of 0 and 1.
We know that chains are different with each other. And their length is exactly n. And what’s more, each chain sequence doesn’t contain “101” as a substring.
Could you tell how many chains will YaoYao have at most?

 

Input

There will be multiple test cases in a test data. For each test case, there is only one number n(n<10000). The end of the input is indicated by a -1, which should not be processed as a case.

 

Output

For each test case, only one line with a number indicating the total number of chains YaoYao can have at most of length n. The answer should be print after module 9997.

 

Sample Input

34-1

 

Sample Output

712
Hint
We can see when the length equals to 4. We can have those chains:0000,0001,0010,00110100,0110,0111,10001001,1100,1110,1111
解题思路：简单递推题
dp1[i] : 为满足不出现101的前提下最后一位是0的方案数
dp2[i] : 为满足不出现101的前提下最后一位是1的方案数
dp[i] : 为总的方案数
#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <string>#include <vector>#include <deque>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <algorithm>#include <functional>using namespace std;const int maxn = 10010;const int mod  = 9997;int dp1[maxn], dp2[maxn], dp[maxn];int main() {    //freopen("aa.in", "r", stdin);    dp1[1] = 1; dp1[2] = 2;    dp2[1] = 1; dp2[2] = 2;    dp[1] = dp1[1] + dp2[1];    dp[2] = dp1[2] + dp2[2];    for(int i = 3; i < 10000; ++i) {        dp1[i] = (dp1[i-1] + dp2[i-1]) % mod;        dp2[i] = (dp1[i-2] + dp2[i-1]) % mod;        dp[i] = (dp1[i] + dp2[i]) % mod;    }    int n;    while(scanf("%d", &n) != EOF) {        if(n < 0) break;        printf("%d\n", dp[n]);    }    return 0;}