HDU 3485 Count 101

最新推荐文章于 2018-08-16 18:36:00 发布

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本文探讨了一道算法题目，要求计算长度为n的不同二进制链的最大数量，这些链中不能包含子串“101”。通过使用动态规划的方法解决了这一问题，并提供了完整的C++代码实现。

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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

3
4
-1

Sample Output

7
12
HintWe can see when the length equals to 4. We can have those chains:
0000,0001,0010,0011
0100,0110,0111,1000
1001,1100,1110,1111

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;
}