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];

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<cstring>
 #define mp make_pair
 #define pb push_back
 #define first fi
 #define second se
 #define pw(x) (1ll << (x))
 #define sz(x) ((int)(x).size())
 #define all(x) (x).begin(),(x).end()
 #define rep(i,l,r) for(int i=(l);i<(r);i++)
 #define per(i,r,l) for(int i=(r);i>=(l);i--)
 #define FOR(i,l,r) for(int i=(l);i<=(r);i++)
 #define eps 1e-9
 #define PIE acos(-1)
 #define cl(a,b) memset(a,b,sizeof(a))
 #define fastio ios::sync_with_stdio(false);cin.tie(0);
 #define lson l , mid , ls
 #define rson mid + 1 , r , rs
 #define ls (rt<<1)
 #define rs (ls|1)
 #define INF 0x3f3f3f3f
 #define LINF 0x3f3f3f3f3f3f3f3f
 #define freopen freopen("in.txt","r",stdin);
 #define cfin ifstream cin("in.txt");
 #define lowbit(x) (x&(-x))
 #define sqr(a) a*a
 #define ll long long
 #define ull unsigned long long
 #define vi vector<int>
 #define pii pair<int, int>
 #define dd(x) cout << #x << " = " << (x) << ", "
 #define de(x) cout << #x << " = " << (x) << "\n"
 #define endl "\n"
 using namespace std;
 //**********************************
 const int maxn=;
 int n;
 ll dp[][];
 //**********************************
 void Init()
 {
     cl(dp,);
     dp[][]=dp[][]=dp[][]=dp[][]=;dp[][]=;
     FOR(i,,){
         dp[i][]=dp[i-][]+dp[i-][]+dp[i-][];
         dp[i][]=dp[i-][];
         dp[i][]=dp[i-][];
 //        dd(i);dd(dp[i][0]);dd(dp[i][1]);de(dp[i][2]);
     }
 }
 //**********************************
 int main()
 {
     Init();
     while(cin>>n,n){
         ll ans=;
         ans=dp[n][]+dp[n][]+dp[n][];
         printf("%lld\n",(1ll<<n)-ans);
     }
     return ;
 }

第二种实现方法

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

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

     /*************************************************************************
         > File Name: a.cpp
         > Author: QWX
         > Mail:
         > Created Time: Sun 30 Sep 2018 04:01:42 AM PDT
      ************************************************************************/

     //{{{ #include
 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<set>
 #include<string>
 #include<cstring>
 #include<complex>
 #include<bits/stdc++.h>
 #define mp make_pair
 #define pb push_back
 #define first fi
 #define second se
 #define pw(x) (1ll << (x))
 #define sz(x) ((int)(x).size())
 #define all(x) (x).begin(),(x).end()
 #define rep(i,l,r) for(int i=(l);i<(r);i++)
 #define per(i,r,l) for(int i=(r);i>=(l);i--)
 #define FOR(i,l,r) for(int i=(l);i<=(r);i++)
 #define eps 1e-9
 #define PIE acos(-1)
 #define cl(a,b) memset(a,b,sizeof(a))
 #define fastio ios::sync_with_stdio(false);cin.tie(0);
 #define lson l , mid , ls
 #define rson mid + 1 , r , rs
 #define ls (rt<<1)
 #define rs (ls|1)
 #define INF 0x3f3f3f3f
 #define LINF 0x3f3f3f3f3f3f3f3f
 #define ll long long
 #define ull unsigned long long
 #define dd(x) cout << #x << " = " << (x) << ","
 #define de(x) cout << #x << " = " << (x) << "\n"
 #define endl "\n"
     using namespace std;
     //}}}

 ll dp[][];
 int n;

 int main()
 {
     dp[][]=dp[][]=;
     FOR(i,,){
         FOR(j,,)dp[i+j][]+=dp[i][];
         FOR(j,,)dp[i+j][]+=dp[i][];
     }
     while(cin>>n,n)cout<<(1ll<<n)-dp[n][]-dp[n][]<<endl;
     return ;
 }

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];

 /*************************************************************************
     > File Name: j.cpp
     > Author: QWX
     > Mail:
     > Created Time: Fri 05 Oct 2018 08:58:42 PM PDT
  ************************************************************************/

 //{{{ #include
 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<set>
 #include<string>
 #include<cstring>
 #include<complex>
 //#include<bits/stdc++.h>
 #define mp make_pair
 #define pb push_back
 #define first fi
 #define second se
 #define vi vector<int>
 #define pii pair<int,int>
 #define pw(x) (1ll << (x))
 #define sz(x) ((int)(x).size())
 #define all(x) (x).begin(),(x).end()
 #define rep(i,l,r) for(int i=(l);i<(r);i++)
 #define per(i,r,l) for(int i=(r);i>=(l);i--)
 #define FOR(i,l,r) for(int i=(l);i<=(r);i++)
 #define eps 1e-9
 #define PIE acos(-1)
 #define cl(a,b) memset(a,b,sizeof(a))
 #define fastio ios::sync_with_stdio(false);cin.tie(0);
 #define lson l , mid , ls
 #define rson mid + 1 , r , rs
 #define ls (rt<<1)
 #define rs (ls|1)
 #define INF 0x3f3f3f3f
 #define LINF 0x3f3f3f3f3f3f3f3f
 #define ll long long
 #define ull unsigned long long
 #define dd(x) cout << #x << " = " << (x) << ","
 #define de(x) cout << #x << " = " << (x) << "\n"
 #define endl "\n"
 using namespace std;
 //}}}

 int n;
 const int maxn=5e5+;
 int dp[maxn][];
 const int mod=1e9+;
 inline int add(int a,int b){a+=b;if(a>mod)a-=mod;return a;}

 int main()
 {
     int n,a,b;
     cin>>n>>a>>b;
     cl(dp,);
     dp[][]=dp[][]=;
     FOR(i,,n){
         FOR(j,,a)dp[i+j][]=add(dp[i+j][],dp[i][]);
         FOR(j,,b)dp[i+j][]=add(dp[i+j][],dp[i][]);
     }
     cout<<add(dp[n][],dp[n][])<<endl;
     return ;
 }