nyoj 01串（dp）

开始做DP。。

第n位为0时，dp[n] = dp[n-1]

第n位为1时，dp[n] =dp[n-2]

得公式dp[n] = dp[n-1] + dp[n-2];

 1 #include<iostream>

 2 #include<cstdio>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<string>

 6 #include<queue>

 7 #include<algorithm>

 8 #include<map>

 9 #include<iomanip>

10 #include<climits>

11 #include<string.h>

12 #include<cmath>

13 #include<stdlib.h>

14 #include<vector>

15 #include<stack>

16 #include<set>

17 #define INF 1e7

18 #define MAXN 100010

19 #define maxn 1000010

20 #define Mod 1000007

21 #define N 1010

22 using namespace std;

23 typedef long long LL;

24 

25 int T, n;

26 int dp[111];

27 

28 void run()

29 {

30     cin >> n;

31     memset(dp, 0, sizeof(dp));

32     dp[2] = 3, dp[3] = 5;

33     for (int i = 4; i < 110; ++i)

34         dp[i] = dp[i - 1] + dp[i - 2];

35     cout << dp[n] << endl;

36 }

37 

38 int main()

39 {

40     cin >> T;

41     while (T--)

42         run();

43     //system("pause");

44     return 0;

45 }