[20180901]四校聯考
阿新 • • 發佈：2018-09-06
info calc == lse void getchar 轉載 get 官方
T1、數列(number)
Solution
首先我們考慮形如\(\frac{n^2}{2}\)的數，顯然n個這樣的數會提供\(\frac{n(n-1)}{2}\)對。
把k看作是幾個形如\(\frac{n(n-1)}{2}\)的和，從大到小貪心加。
要保證任意兩個不同的數的和不是完全平方數，暴力構造一下就可以了。
#include<iostream> #include<cstdio> #include<algorithm> typedef long long ll; ll K,num[100005],a[155]; ll ans,print[155]; ll l,r,res,pos=0; const ll nn[105]={0,1,3,5,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49,51,53,55,57,59,61,63,65,67}; int main(){ freopen("number.in","r",stdin); freopen("number.out","w",stdout); scanf("%lld",&K); if(K<100000){ if(K<=6){ printf("%lld\n",K+1);printf("1 "); for(int i=1;i<=K;i++) printf("3 "); } else{ printf("%d\n",K-1); printf("2 2 2 2 ");printf("1 "); for(int i=1;i<=K-6;i++) printf("3 "); } return 0; } for(int i=1;i<=100000;i++) num[i]=1LL*i*(i-1)/2; for(int i=1;i<=100;i++) a[i]=2LL*nn[i]*nn[i]; while(K!=0){ ++pos; l=2;r=100000; while(l<=r){ ll mid=(l+r)>>1; if(num[mid]<=K) res=mid,l=mid+1; else r=mid-1; } print[pos]=res; K-=num[res];ans+=res; } printf("%lld\n",ans); for(register int i=1;i<=pos;i++) for(register int j=1;j<=print[i];j++) printf("%lld ",a[i]); return 0; }
T2、散步(walk)
Solution
對於15分，直接用map來記錄就可以直接判斷了。
官方題解
/*15 points #include<iostream> #include<cstdio> #include<algorithm> #include<map> #define MN 100005 using namespace std; int x,y,n,k,ans; char s[MN]; std::map<std::pair<int,int>,bool> mp; std::map<std::pair<int,int>,bool> Mp; int main(){ freopen("walk.in","r",stdin); freopen("walk.out","w",stdout); scanf("%d%d",&n,&k); scanf("%s",s+1); if(k!=1) return 0*puts("orz!"); x=0;y=0;mp[make_pair(0,0)]=true; for(int i=1;i<=n;i++){ if(s[i]=='W'){x-=1;} if(s[i]=='E'){x+=1;} if(s[i]=='S'){y-=1;} if(s[i]=='N'){y+=1;} mp[make_pair(x,y)]=true; if(mp[make_pair(x-1,y)]&&mp[make_pair(x-1,y-1)]&&mp[make_pair(x,y-1)]&&Mp[make_pair(x-1,y-1)]!=true) Mp[make_pair(x-1,y-1)]=true,ans++; if(mp[make_pair(x-1,y)]&&mp[make_pair(x-1,y+1)]&&mp[make_pair(x,y+1)]&&Mp[make_pair(x-1,y)]!=true) Mp[make_pair(x-1,y)]=true,ans++; if(mp[make_pair(x+1,y)]&&mp[make_pair(x+1,y-1)]&&mp[make_pair(x,y-1)]&&Mp[make_pair(x,y-1)]!=true) Mp[make_pair(x,y-1)]=true,ans++; if(mp[make_pair(x+1,y)]&&mp[make_pair(x+1,y+1)]&&mp[make_pair(x,y+1)]&&Mp[make_pair(x,y)]!=true) Mp[make_pair(x,y)]=true,ans++; } printf("%d\n",ans); }*/ #include<bits/stdc++.h> #define ll long long #define F(i,a,b) for(i=a;i<=b;i++) inline int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();} return x*f; } #define MN 100005 #define depail std::deque<std::pair<int,int> > #define MAP std::map<std::pair<int,int>,depail > int n,k,a[MN],b[MN],p,q; char s[MN];MAP mp;ll ans; int calc(int tp,depail a,depail b,depail c,depail d){ for(depail::iterator i=a.begin();i!=a.end();i++) i->first+=tp,i->second+=tp; for(depail::iterator i=b.begin();i!=b.end();i++) i->first+=tp,i->second+=tp; int res=0; #define f(a) (a.empty()?-1e9:a.front().second) #define G(a) a.front().first #define H(a) if(f(a)==r) a.pop_front(); int l=-1e9; while(1){ int r=std::min(std::min(f(a),f(b)),std::min(f(c),f(d))); if(r==-1e9) break; l=std::max(l,std::max(std::max(G(a),G(b)),std::max(G(c),G(d)))); if(l<=r) res+=r-l+1,l=r+1; H(a);H(b);H(c);H(d); } return res; } int main(){ freopen("walk.in","r",stdin); freopen("walk.out","w",stdout); scanf("%d%d",&n,&k); scanf("%s",s+1); register int i,x=0,y=0; F(i,1,n){ if(s[i]=='W'){x-=1;} if(s[i]=='E'){x+=1;} if(s[i]=='S'){y-=1;} if(s[i]=='N'){y+=1;} a[i]=x;b[i]=y; } if(a[n]<0) F(i,1,n) a[i]=-a[i]; if(b[n]<0) F(i,1,n) b[i]=-b[i]; if(a[n]==0) F(i,1,n) std::swap(a[i],b[i]); p=a[n]?a[n]:-1e9;q=b[n]; if(a[n]==0) F(i,0,n) mp[std::make_pair(a[i],b[i])].push_back(std::make_pair(0,0)); else F(i,0,n){ int d=a[i]/p; if(a[i]-p*d<0) d--; mp[std::make_pair(a[i]-p*d,b[i]-q*d)].push_back(std::make_pair(d,d+k-1)); // x=a[i]%p;y=b[i]-(a[i]/p)*q; // mp[std::make_pair(x,y)].push_back(std::make_pair(a[i]/p,a[i]/p+k-1)); } for(MAP::iterator i=mp.begin();i!=mp.end();++i) sort(i->second.begin(),i->second.end()); for(MAP::iterator i=mp.begin();i!=mp.end();++i){ x=i -> first.first,y=i -> first.second; #define solve(d,p2,p3,p4) {if(mp.count(p2)&&mp.count(p3)&&mp.count(p4)) ans+=calc(d,i->second,mp[p2],mp[p3],mp[p4]);} if(x!=p-1) solve(0,std::make_pair(x,y+1),std::make_pair(x+1,y),std::make_pair(x+1,y+1)) else solve(1,std::make_pair(x,y+1),std::make_pair(0,y-q),std::make_pair(0,y-q+1)) } printf("%lld",ans); return 0; }
T3、考古(archaeology)
Solution
把操作反過來，就可以看成是，每次會有一段地面下降，問最後每一段地面會降至那一層。
用樹狀數組/線段樹來維護下降過程中每一段的(x,y)值，尋找下降的區間可以直接在樹狀數組/線段樹上二分。
#include<bits/stdc++.h> using namespace std; inline int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();} return x*f; } #define MN 200005 #define ll long long #define lowbit(a) (a&-a) int n,q,x[MN],l[MN]; ll xpos[MN],ypos[MN],pos,X,Y; bool d[MN]; void Cx(int x,int val){for(;x<=n;x+=lowbit(x))xpos[x]+=val;} void Cy(int x,int val){for(;x<=n;x+=lowbit(x))ypos[x]+=val;} ll G(int x){ll res=0;for(;x;x-=lowbit(x)) res+=ypos[x];return res;} int main(){ freopen("archaeology.in","r",stdin); freopen("archaeology.out","w",stdout); n=read();q=read(); register int i,j; for(i=1;i<=q;++i) x[i]=read(),d[i]=read()==1,l[i]=read(); for(i=2;i<=n;++i) Cx(i,1); for(i=q;i;--i){ if(d[i]){ pos=X=Y=0; for(j=17;~j;--j)if((pos|1<<j)<=n&&X+xpos[pos|1<<j]-Y-ypos[pos|1<<j]<x[i]) pos|=1<<j,X+=xpos[pos],Y+=ypos[pos]; Cx(1,-l[i]),Cy(1,-l[i]); Cx(pos+1,l[i]),Cy(pos+1,l[i]); } else{ pos=X=Y=0; for(j=17;~j;--j)if((pos|1<<j)<=n&&X+xpos[pos|1<<j]+Y+ypos[pos|1<<j]<x[i]) pos|=1<<j,X+=xpos[pos],Y+=ypos[pos]; Cx(pos+1,l[i]),Cy(pos+1,-l[i]); } } for(i=1;i<=n;i++) printf("%lld\n",-G(i)); return 0; }
Blog來自PaperCloud，未經允許，請勿轉載，TKS！[]
[20180901]四校聯考