dp[i][j]长为i,最高位j,没有lead0的个数。然后大到小。如果dp[i][0]<n就n中减去。具体看代码。
#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <vector> #include <iomanip> #include <cstring> #include <map> #include <queue> #include <set> #include <cassert> #include <stack> #include <bitset> //#include <unordered_set> #define mkp make_pair #define err cout<<"here"<<endl using namespace std; const double EPS=1e-8; typedef long long lon; typedef unsigned long long ull; typedef pair<int,int> pii; typedef map<ull,int>::iterator IT; const lon SZ=71,SSZ=200010,APB=18,mod=100000,one=97; const lon INF=0x7FFFFFFF; lon n,dp[SZ][2]; struct nd{ lon to,wt; nd(lon a=0,lon b=0):to(a),wt(b){} }; lon dfs(lon len,lon val,lon lead) { if(!lead&&dp[len][val]!=-1)return dp[len][val]; if(len==1) { return !lead; } lon res=0; if(val==1)res=dfs(len-1,0,0); else res=dfs(len-1,0,lead)+dfs(len-1,1,0); //if(len==3)cout<<dfs(len-1,0,lead)<<endl; if(!lead)dp[len][val]=res; return res; } bool ok=0; void init() { cin>>n; lon lead=1; for(lon i=SZ-1;i>=1;--i) { //cout<<i<<" "<<dfs(i,0,1)<<" "<<n<<endl; if(dfs(i,0,1)<n)lead=0,n-=dfs(i,0,1)+1,cout<<1; else if(!lead)cout<<0; //err; } cout<<endl; } void work() { } void release() { //memset(dp,0,sizeof(dp)); } int main() { std::ios::sync_with_stdio(0); //freopen("d:\\1.txt","r",stdin); memset(dp,-1,sizeof(dp)); //cout<<dfs(3,0,1)<<endl; for(lon i=1;i<SZ;++i) { dfs(i,1,0),dfs(i,0,1); //cout<<i<<" "<<dfs(i,0,1)<<" "<<dfs(i,1,0)<<endl; } lon casenum; cin>>casenum; //cout<<casenum<<endl; for(int time=1;time<=casenum;++time) //for(int time=1;;++time) { cout<<"Case "<<time<<": "; init(); work(); release(); } return 0; }
原文:https://www.cnblogs.com/gaudar/p/10862576.html