前言：单独对题面描述的评分->
【题解】把相邻长度为2的子串两两连边，跑欧拉路
/*明天再写，先贴一份方老师代码压压惊*/
#include<map> #include<stack> #include<queue> #include<cstdio> #include<string> #include<vector> #include<cstring> #include<complex> #include<iostream> #include<assert.h> #include<algorithm> using namespace std; #define inf 1001001001 #define infll 1001001001001001001LL #define FOR0(i,n) for(int (i)=0;(i)<(n);++(i)) #define FOR1(i,n) for(int (i)=1;(i)<=(n);++(i)) #define ll long long #define dbg(vari) cerr<<#vari<<" = "<<(vari)<<endl #define gmax(a,b) (a)=max((a),(b)) #define gmin(a,b) (a)=min((a),(b)) #define ios0 ios_base::sync_with_stdio(0) #define Ri register int #define gc getchar() #define il inline il int read(){ bool f=true; Ri x=0;char ch; while(!isdigit(ch=gc))if(ch=='-')f=false; while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=gc;} return f?x:-x; } #define gi read() #define FO(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout); int hash(char a,char b){ return (int)a*333+b; } char A,B; void rhash(int a){ B=a%333; A=(a-B)/333; } struct edge{ int to,next; }e[666666]; int cnt,last[666666],dgr[666666]; void insert(int a,int b){ e[++cnt]=(edge){b,last[a]};last[a]=cnt;++dgr[a];++dgr[b]; } int T,mark[666666],ans[666666],_ans=0; void dfs(int v){ rhash(v); ans[++_ans]=v; for(int i=last[v];i;i=e[i].next){ if(!mark[i]){ mark[i]=true; dfs(e[i].to); } } } int main(){ FO(recover); T=gi; while(T--){ int n=gi;_ans=0; memset(last,0,sizeof(last)); memset(dgr,0,sizeof(dgr)); cnt=0;int source=0; FOR1(i,n){ char str[6]; scanf("%s",str+1); if(!source)source=hash(str[1],str[2]); insert(hash(str[1],str[2]),hash(str[2],str[3])); } int oddcnt=0; FOR1(i,hash('z','z')){ oddcnt+=dgr[i]&1; } if(oddcnt>2||oddcnt==1)puts("NO"); else{ if(oddcnt){ FOR1(i,hash('z','z')){ if(dgr[i]&1){ dfs(i);break; } } }else{ dfs(source); } } FOR1(i,_ans){ rhash(ans[i]); if(i==1){putchar(A),putchar(B);} else{putchar(B);} } puts(""); } }
【题解】
题意相当于是给定相邻两两的大小关系，求方案数
那么可以得到一个dp方程dp[i][j] 表示1-i的排列数为j的方案数
那么
时
否则
这样看起来是的，很显然可以应用前缀和优化到
#include<map> #include<stack> #include<queue> #include<cstdio> #include<string> #include<vector> #include<cstring> #include<complex> #include<iostream> #include<assert.h> #include<algorithm> using namespace std; #define pb push_back #define inf 1001001001 #define infll 1001001001001001001LL #define FOR0(i,n) for(int (i)=0;(i)<(n);++(i)) #define FOR1(i,n) for(int (i)=1;(i)<=(n);++(i)) #define mp make_pair #define pii pair<int,int> #define ll long long #define ld double #define vi vector<int> #define SZ(x) ((int)((x).size())) #define fi first #define se second #define RI(n) int (n); scanf("%d",&(n)); #define RI2(n,m) int (n),(m); scanf("%d %d",&(n),&(m)); #define RI3(n,m,k) int (n),(m),(k); scanf("%d %d %d",&(n),&(m),&(k)); template<typename T,typename TT> ostream& operator<<(ostream &s,pair<T,TT> t) {return s<<"("<<t.first<<","<<t.second<<")";} template<typename T> ostream& operator<<(ostream &s,vector<T> t){FOR0(i,sz(t))s<<t[i]<<" ";return s; } #define dbg(vari) cerr<<#vari<<" = "<<(vari)<<endl #define all(t) t.begin(),t.end() #define FEACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++) #define TESTS RI(testow)while(testow--) #define FORZ(i,a,b) for(int (i)=(a);(i)<=(b);++i) #define FORD(i,a,b) for(int (i)=(a); (i)>=(b);--i) #define gmax(a,b) (a)=max((a),(b)) #define gmin(a,b) (a)=min((a),(b)) #define ios0 ios_base::sync_with_stdio(0) using namespace std; #define Ri register int #define gc getchar() #define il inline il int read(){ bool f=true; Ri x=0;char ch; while(!isdigit(ch=gc))if(ch=='-')f=false; while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=gc;} return f?x:-x; } #define gi read() #define FO(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout); int a[2333],s[2333]; int b[2333],ans,f[2333][2333]={{0,1}}; int main(){ FO(good); TESTS{ RI2(n,m); memset(b,0,sizeof(b)); FOR1(i,m)s[i]=gi,b[s[i]]=true; FOR1(i,n-1){ int t=0; if(b[i]) FOR1(k,i+1){ f[i][k]=t; t=(t+f[i-1][k])%1000000007; }else FORD(k,i+1,1){ t=(t+f[i-1][k])%1000000007; f[i][k]=t; } } int ans=0; FOR1(i,n)ans=(ans+f[n-1][i])%1000000007; cout<<ans<<endl; } return 0; }
T3
题面
输入n,s
当s=1时，是http://codeforces.com/contest/690/problem/A1
当s=0时，是http://codeforces.com/contest/690/problem/A2
【题解】
首先s=1的情况比较好做，答案就是
当s=0时，我们可以发现，当时，ZY一个钻石都不要出，就能活命
那么当对于多出来的2k人，可以沿用s=1的策略
如果n是一个奇数，更加简单啊，答案是
#include<map> #include<stack> #include<queue> #include<cstdio> #include<string> #include<vector> #include<cstring> #include<complex> #include<iostream> #include<assert.h> #include<algorithm> using namespace std; #define pb push_back #define inf 1001001001 #define infll 1001001001001001001LL #define FOR0(i,n) for(int (i)=0;(i)<(n);++(i)) #define FOR1(i,n) for(int (i)=1;(i)<=(n);++(i)) #define mp make_pair #define pii pair<int,int> #define ll long long #define ld double #define vi vector<int> #define SZ(x) ((int)((x).size())) #define fi first #define se second #define RI(n) int (n); scanf("%d",&(n)); #define RI2(n,m) int (n),(m); scanf("%d %d",&(n),&(m)); #define RI3(n,m,k) int (n),(m),(k); scanf("%d %d %d",&(n),&(m),&(k)); template<typename T,typename TT> ostream& operator<<(ostream &s,pair<T,TT> t) {return s<<"("<<t.first<<","<<t.second<<")";} template<typename T> ostream& operator<<(ostream &s,vector<T> t){FOR0(i,sz(t))s<<t[i]<<" ";return s; } #define dbg(vari) cerr<<#vari<<" = "<<(vari)<<endl #define all(t) t.begin(),t.end() #define FEACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++) #define TESTS RI(testow)while(testow--) #define FORZ(i,a,b) for(int (i)=(a);(i)<=(b);++i) #define FORD(i,a,b) for(int (i)=(a); (i)>=(b);--i) #define gmax(a,b) (a)=max((a),(b)) #define gmin(a,b) (a)=min((a),(b)) #define ios0 ios_base::sync_with_stdio(0) using namespace std; #define Ri register int #define gc getchar() #define il inline il int read(){ bool f=true; Ri x=0;char ch; while(!isdigit(ch=gc))if(ch=='-')f=false; while(isdigit(ch)){x=(x<<1)+(x<<3)+ch-'0';ch=gc;} return f?x:-x; } #define gi read() #define FO(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout); int main() { FO(distribute); TESTS{ int n,s; n=gi;s=gi; if(s==1)printf("%d ",(n+1)/2); else{ if(n&1){ printf("%d ",n/2); continue; } n/=2; int d=0; while((1<<d)<=n) ++d; --d; n-=(1<<d); printf("%d ",n); } } return 0; }