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递推,求至少连续放置三个U的危险组合

UVA580-Critical Mass

题意

有两种方块,L和U,有至少三个连续的U称为危险组合,问有多少个危险组合

solution:

至少这个概念比较难求 ,所以转化为(1ll<<n)-安全组合

dp[n][i]表示前n个数里以i个U结尾的个数

递推方程

dp[i][0]=dp[i-1][0]+dp[i-1][1]+dp[i-1][2];
dp[i][1]=dp[i-1][0];
dp[i][2]=dp[i-1][1];

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<vector>
 5 #include<cmath>
 6 #include<queue>
 7 #include<cstring>
 8 #define mp make_pair
 9 #define pb push_back
10 #define first fi
11 #define second se
12 #define pw(x) (1ll << (x))
13 #define sz(x) ((int)(x).size())
14 #define all(x) (x).begin(),(x).end()
15 #define rep(i,l,r) for(int i=(l);i<(r);i++)
16 #define per(i,r,l) for(int i=(r);i>=(l);i--)
17 #define FOR(i,l,r) for(int i=(l);i<=(r);i++)
18 #define eps 1e-9
19 #define PIE acos(-1)
20 #define cl(a,b) memset(a,b,sizeof(a))
21 #define fastio ios::sync_with_stdio(false);cin.tie(0);
22 #define lson l , mid , ls
23 #define rson mid + 1 , r , rs
24 #define ls (rt<<1)
25 #define rs (ls|1)
26 #define INF 0x3f3f3f3f
27 #define LINF 0x3f3f3f3f3f3f3f3f
28 #define freopen freopen("in.txt","r",stdin);
29 #define cfin ifstream cin("in.txt");
30 #define lowbit(x) (x&(-x))
31 #define sqr(a) a*a
32 #define ll long long
33 #define ull unsigned long long
34 #define vi vector<int>
35 #define pii pair<int, int>
36 #define dd(x) cout << #x << " = " << (x) << ", "
37 #define de(x) cout << #x << " = " << (x) << "\n"
38 #define endl "\n"
39 using namespace std;
40 //**********************************
41 const int maxn=30;
42 int n;
43 ll dp[35][3];
44 //**********************************
45 void Init()
46 {
47     cl(dp,0);
48     dp[1][0]=dp[2][2]=dp[1][1]=dp[2][1]=1;dp[2][0]=2;
49     FOR(i,3,30){
50         dp[i][0]=dp[i-1][0]+dp[i-1][1]+dp[i-1][2];
51         dp[i][1]=dp[i-1][0];
52         dp[i][2]=dp[i-1][1];
53 //        dd(i);dd(dp[i][0]);dd(dp[i][1]);de(dp[i][2]);
54     }
55 }
56 //**********************************
57 int main()
58 {
59     Init();
60     while(cin>>n,n){
61         ll ans=0;
62         ans=dp[n][0]+dp[n][1]+dp[n][2];
63         printf("%lld\n",(1ll<<n)-ans);
64     }
65     return 0;
66 }
View Code

 第二种实现方法

dp[i+j][1]+=dp[i][0];

dp[i+j][0]+=dp[i][1];

 1     /*************************************************************************
 2         > File Name: a.cpp
 3         > Author: QWX
 4         > Mail: 
 5         > Created Time: Sun 30 Sep 2018 04:01:42 AM PDT
 6      ************************************************************************/
 7 
 8     //{{{ #include
 9 #include<iostream>
10 #include<cstdio>
11 #include<algorithm>
12 #include<vector>
13 #include<cmath>
14 #include<queue>
15 #include<map>
16 #include<set>
17 #include<string>
18 #include<cstring>
19 #include<complex>
20 #include<bits/stdc++.h>
21 #define mp make_pair
22 #define pb push_back
23 #define first fi
24 #define second se
25 #define pw(x) (1ll << (x))
26 #define sz(x) ((int)(x).size())
27 #define all(x) (x).begin(),(x).end()
28 #define rep(i,l,r) for(int i=(l);i<(r);i++)
29 #define per(i,r,l) for(int i=(r);i>=(l);i--)
30 #define FOR(i,l,r) for(int i=(l);i<=(r);i++)
31 #define eps 1e-9
32 #define PIE acos(-1)
33 #define cl(a,b) memset(a,b,sizeof(a))
34 #define fastio ios::sync_with_stdio(false);cin.tie(0);
35 #define lson l , mid , ls
36 #define rson mid + 1 , r , rs
37 #define ls (rt<<1)
38 #define rs (ls|1)
39 #define INF 0x3f3f3f3f
40 #define LINF 0x3f3f3f3f3f3f3f3f
41 #define ll long long
42 #define ull unsigned long long
43 #define dd(x) cout << #x << " = " << (x) << "," 
44 #define de(x) cout << #x << " = " << (x) << "\n" 
45 #define endl "\n"
46     using namespace std;
47     //}}}
48 
49 
50 
51 ll dp[100][2];
52 int n;
53 
54 int main()
55 {
56     dp[0][0]=dp[0][1]=1;
57     FOR(i,0,30){
58         FOR(j,1,2)dp[i+j][0]+=dp[i][1];
59         FOR(j,1,30)dp[i+j][1]+=dp[i][0];
60     }
61     while(cin>>n,n)cout<<(1ll<<n)-dp[n][0]-dp[n][1]<<endl;
62     return 0;
63 }
View Code

 

 

类似题目

The Debut Album

题意:

给你一个串的长度n,串由1,2组成,1连续不超过a个,2连续不超过b个,问有多少种这样的串,对1e9+7取模。

solution:

递推方程

往后补值:dp[i+j][1]+=dp[i][0];

     dp[i+j][0]+=dp[i][1];

向前取值:dp[i][1]+=dp[j][0];

     dp[i][0]+=dp[j][1];

 

 

 1 /*************************************************************************
 2     > File Name: j.cpp
 3     > Author: QWX
 4     > Mail: 
 5     > Created Time: Fri 05 Oct 2018 08:58:42 PM PDT
 6  ************************************************************************/
 7 
 8 
 9 //{{{ #include
10 #include<iostream>
11 #include<cstdio>
12 #include<algorithm>
13 #include<vector>
14 #include<cmath>
15 #include<queue>
16 #include<map>
17 #include<set>
18 #include<string>
19 #include<cstring>
20 #include<complex>
21 //#include<bits/stdc++.h>
22 #define mp make_pair
23 #define pb push_back
24 #define first fi
25 #define second se
26 #define vi vector<int>
27 #define pii pair<int,int>
28 #define pw(x) (1ll << (x))
29 #define sz(x) ((int)(x).size())
30 #define all(x) (x).begin(),(x).end()
31 #define rep(i,l,r) for(int i=(l);i<(r);i++)
32 #define per(i,r,l) for(int i=(r);i>=(l);i--)
33 #define FOR(i,l,r) for(int i=(l);i<=(r);i++)
34 #define eps 1e-9
35 #define PIE acos(-1)
36 #define cl(a,b) memset(a,b,sizeof(a))
37 #define fastio ios::sync_with_stdio(false);cin.tie(0);
38 #define lson l , mid , ls
39 #define rson mid + 1 , r , rs
40 #define ls (rt<<1)
41 #define rs (ls|1)
42 #define INF 0x3f3f3f3f
43 #define LINF 0x3f3f3f3f3f3f3f3f
44 #define ll long long
45 #define ull unsigned long long
46 #define dd(x) cout << #x << " = " << (x) << "," 
47 #define de(x) cout << #x << " = " << (x) << "\n" 
48 #define endl "\n"
49 using namespace std;
50 //}}}
51 
52 
53 
54 int n;
55 const int maxn=5e5+7;
56 int dp[maxn][2];
57 const int mod=1e9+7;
58 inline int add(int a,int b){a+=b;if(a>mod)a-=mod;return a;}
59 
60 int main()
61 {
62     int n,a,b;
63     cin>>n>>a>>b;
64     cl(dp,0);
65     dp[0][0]=dp[0][1]=1;
66     FOR(i,0,n){
67         FOR(j,1,a)dp[i+j][0]=add(dp[i+j][0],dp[i][1]);
68         FOR(j,1,b)dp[i+j][1]=add(dp[i+j][1],dp[i][0]);
69     }
70     cout<<add(dp[n][0],dp[n][1])<<endl;
71     return 0;
72 }
View Code

 

 

 

 

 

posted @ 2018-09-21 20:31  牛奶加咖啡~  阅读(144)  评论(0编辑  收藏  举报