A. Count The Ones
$ans=b-c+1$。
#include <stdio.h> using namespace std ; int a , b , c ; void solve () { printf ( "%d\n" , 1 + b - c ) ; } int main () { while ( ~scanf ( "%d%d%d" , &a , &b , &c ) ) solve () ; return 0 ; }
B. Craters
求出凸包,然後列舉凸包上兩個點,對第三個點旋轉卡殼。因為隨機資料凸包期望點數為$O(\sqrt{n})$,故時間複雜度為$O(n\log n)$。
#include<cstdio> #include<cmath> #include<algorithm> #include<set> using namespace std; typedef long long ll; typedef pair<int,int>PI; const int N=200010; int n,m,_,i,j,k,x,y; set<PI>T; ll ans=-1; struct P{ int x,y; P(){} P(int _x,int _y){x=_x,y=_y;} P operator-(P b){return P(x-b.x,y-b.y);} bool operator<(const P&p)const{ if(x!=p.x)return x<p.x; return y<p.y; } void write(){ printf("%d %d\n",x,y); } }a[N],b[N<<1],ans0,ans1,ans2; inline ll cross(P a,P b){ return 1LL*a.x*b.y-1LL*a.y*b.x; } inline ll vect(P p,P p1,P p2){ return 1LL*(p1.x-p.x)*(p2.y-p.y)-1LL*(p1.y-p.y)*(p2.x-p.x); } int convexhull(P*p,int n,P*q){ int i,k,m; sort(p,p+n); m=0; for(i=0;i<n;q[m++]=p[i++])while(m>1&&vect(q[m-2],q[m-1],p[i])<=0)m--; k=m; for(i=n-2;i>=0;q[m++]=p[i--])while(m>k&&vect(q[m-2],q[m-1],p[i])<=0)m--; return --m; } inline void work(P a,P b,P c){ ll now=cross(a,b)+cross(b,c)+cross(c,a); if(now<0)now=-now; if(now>ans)ans=now,ans0=a,ans1=b,ans2=c; } inline ll area(P a,P b,P c){ ll now=cross(a,b)+cross(b,c)+cross(c,a); if(now<0)now=-now; return now; } int main(){ scanf("%d",&n); if(n<200){ for(i=0;i<n;i++)scanf("%d%d",&a[i].x,&a[i].y); for(i=0;i<n;i++)for(j=0;j<i;j++)for(k=0;k<j;k++){ work(a[i],a[j],a[k]); } ans0.write(); ans1.write(); ans2.write(); return 0; } _=n; n=0; while(_--){ scanf("%d%d",&x,&y); if(T.find(PI(x,y))!=T.end())continue; a[n++]=P(x,y); T.insert(PI(x,y)); } m=convexhull(a,n,b); for(i=0;i<m;i++)b[i+m]=b[i]; for(i=0;i<m;i++){ for(j=i+1,k=j+1;j<i+m;j++){ if(k<=j)k=j+1; while(k+1<i+m&&area(b[i],b[j],b[k])<=area(b[i],b[j],b[k+1]))k++; work(b[i],b[j],b[k]); } } ans0.write(); ans1.write(); ans2.write(); }
C. MSTrikes back!
記錄最後$5$個點連通性的最小表示然後DP,注意到邊權的週期不會太大,所以可以直接跳過完整的週期。
#include<stdio.h> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<vector> #include<set> #include<map> #include<queue> #include<time.h> #include<assert.h> #include<iostream> using namespace std; typedef long long LL; typedef pair<int,int>pi; const int Ms=3125; const LL Inf=1LL<<60; const int LIM=100; int n,seed; LL in[LIM+1]; LL f[LIM+1][Ms]; LL dp[2][Ms]; void init(int cs,LL val){ for(int i=0;i<Ms;i++)dp[cs][i]=val; } int g[Ms][(1<<5)+1]; void decode(int mask,int *a){ for(int i=0;i<5;i++){ a[i]=mask%5; mask/=5; } } void zuixiao(int *a){ int dui[10]; memset(dui,-1,sizeof dui); int cur=0; for(int i=0;i<5;i++){ int x=a[i]; if(dui[x]<0){ dui[x]=cur++; } a[i]=dui[x]; } } int encode(int *a){ int res=0,tmp=1; for(int i=0;i<5;i++,tmp=tmp*5){ res+=a[i]*tmp; } return res; } bool testt(int *a){ int b[10]; for(int i=0;i<5;i++)b[i]=a[i]; zuixiao(b); for(int i=0;i<5;i++)if(b[i]!=a[i])return 0; return 1; } void pre(){ int a[10],b[10]; for(int mask=0;mask<Ms;mask++){ decode(mask,a); if(!testt(a)){ for(int j=0;j<1<<5;j++)g[mask][j]=-1; continue; } //for(int i=0;i<5;i++)printf("%d ",a[i]);puts(":\n"); for(int j=0;j<1<<5;j++){ for(int i=0;i<5;i++)b[i]=a[i]; b[5]=6; for(int k=0;k<5;k++){ if(j>>k&1){ for(int t=0;t<5;t++){ if(t==k)continue; if(b[t]==b[k])b[t]=b[5]; } b[k]=b[5]; } } bool flag=0; for(int k=1;k<=5;k++)if(b[0]==b[k]){flag=1;break;} if(!flag)g[mask][j]=-1; else{ for(int k=0;k<5;k++)b[k]=b[k+1]; zuixiao(b); //for(int k=0;k<5;k++)printf("%d ",b[k]);puts(""); g[mask][j]=encode(b); } } } } void upd(LL &x,LL y){if(x>y)x=y;} int main(){ pre(); int _;scanf("%d",&_); while(_--){ scanf("%d%d",&n,&seed); int cs=0; init(cs,Inf); dp[cs][0]=0; for(int i=2;i<=n&&i<=LIM;i++){ init(cs^1,Inf); seed=seed*907%2333333; LL bq[10]; int T=seed; in[i]=T; for(int j=0;j<5;j++){ if(i-(5-j)<1)bq[j]=Inf; else { seed=seed*907%2333333; bq[j]=seed^T; } } for(int mask=0;mask<Ms;mask++){ LL w=dp[cs][mask]; if(w==Inf)continue; for(int j=0;j<1<<5;j++){ if(g[mask][j]<0)continue; bool flag=1; for(int k=0;k<5;k++)if((j>>k&1)&&(bq[k]==Inf)){ flag=0;continue; } if(!flag)continue; LL nw=w; for(int k=0;k<5;k++){ if(j>>k&1)nw+=bq[k]; } upd(dp[cs^1][g[mask][j]],nw); } } cs^=1; for(int j=0;j<Ms;j++)f[i][j]=dp[cs][j]; } if(n<=LIM)printf("%lld\n",dp[cs][0]); else{ LL pace,pacew; for(int i=LIM-1;;i--)if(in[i]==in[LIM]){ pace=LIM-i; break; } int st=(n-LIM)/pace*pace+LIM; //printf("st=%d\n",st); for(int i=0;i<Ms;i++){ dp[cs][i]=f[LIM][i]+1LL*(n-LIM)/pace*(f[LIM][i]-f[LIM-pace][i]); } for(int i=st+1;i<=n;i++){ init(cs^1,Inf); seed=seed*907%2333333; LL bq[10]; int T=seed; for(int j=0;j<5;j++){ if(i-(5-j)<1)bq[j]=Inf; else { seed=seed*907%2333333; bq[j]=seed^T; } } for(int mask=0;mask<Ms;mask++){ LL w=dp[cs][mask]; if(w==Inf)continue; for(int j=0;j<1<<5;j++){ if(g[mask][j]<0)continue; bool flag=1; for(int k=0;k<5;k++)if((j>>k&1)&&(bq[k]==Inf)){ flag=0;continue; } if(!flag)continue; LL nw=w; for(int k=0;k<5;k++){ if(j>>k&1)nw+=bq[k]; } upd(dp[cs^1][g[mask][j]],nw); } } cs^=1; } printf("%lld\n",dp[cs][0]); } } return 0; }
D. Skyscrapers
首先用set維護所有已經被摧毀的建築。然後用線段樹維護區間內最容易被向左/向右的衝擊波擊毀的建築即可。時間複雜度$O(n\log n)$。
#include<cstdio> #include<set> #include<algorithm> using namespace std; const int N=100010,M=262150; int n,m,i,x,ans,f[N],g[N]; set<int>T; int vl[M],vr[M],pos[N],p; inline int mergel(int x,int y){ if(!x||!y)return x+y; return f[x]>f[y]?x:y; } inline int merger(int x,int y){ if(!x||!y)return x+y; return g[x]<g[y]?x:y; } void build(int x,int a,int b){ if(a==b){ pos[a]=x; vl[x]=vr[x]=a; return; } int mid=(a+b)>>1; build(x<<1,a,mid); build(x<<1|1,mid+1,b); vl[x]=mergel(vl[x<<1],vl[x<<1|1]); vr[x]=merger(vr[x<<1],vr[x<<1|1]); } void askl(int x,int a,int b,int c,int d){ if(c<=a&&b<=d){ p=mergel(p,vl[x]); return; } int mid=(a+b)>>1; if(c<=mid)askl(x<<1,a,mid,c,d); if(d>mid)askl(x<<1|1,mid+1,b,c,d); } void askr(int x,int a,int b,int c,int d){ if(c<=a&&b<=d){ p=merger(p,vr[x]); return; } int mid=(a+b)>>1; if(c<=mid)askr(x<<1,a,mid,c,d); if(d>mid)askr(x<<1|1,mid+1,b,c,d); } inline void kill(int x){ ans++; T.insert(x); x=pos[x]; vl[x]=vr[x]=0; for(x>>=1;x;x>>=1){ vl[x]=mergel(vl[x<<1],vl[x<<1|1]); vr[x]=merger(vr[x<<1],vr[x<<1|1]); } } inline void workl(int l,int r,int t){ if(l>r)return; while(1){ p=0; askl(1,1,n,l,r); if(!p)return; if(f[p]<t)return; kill(p); } } inline void workr(int l,int r,int t){ if(l>r)return; while(1){ p=0; askr(1,1,n,l,r); if(!p)return; if(g[p]>t)return; kill(p); } } int main(){ scanf("%d",&n); for(i=1;i<=n;i++)scanf("%d",&x),f[i]=i-x,g[i]=i+x; T.insert(0); T.insert(n+1); build(1,1,n); scanf("%d",&m); while(m--){ scanf("%d",&x); ans=0; kill(x); set<int>::iterator j=T.find(x),k; k=j; j--;k++; workl(*j+1,x-1,f[x]); workr(x+1,*k-1,g[x]); printf("%d\n",ans); } /* left: a[x]-a[y]>=x-y y-a[y]>=x-a[x] get max(y-a[y]) right: a[x]-a[y]>=y-x y+a[y]<=x+a[x] get min(y+a[y]) */ }
E. Blackboard
按題意模擬即可。
#include<stdio.h> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<vector> #include<set> #include<map> #include<queue> #include<time.h> #include<assert.h> #include<iostream> using namespace std; typedef long long LL; typedef pair<int,int>pi; int n,ty; int a[103][103]; void rev(){ for(int i=1;i<=n;i++){ for(int j=i;j<=n;j++){ swap(a[i][j],a[j][i]); } } } void solve1(){ int cur=1; for(int i=1;i<=n;i++){ if(i&1){ for(int j=1;j<=n;j++){ a[i][j]=cur++; } } else{ for(int j=n;j>=1;j--){ a[i][j]=cur++; } } } } bool isok(int x,int y){ return x>=1&&x<=n&&y>=1&&y<=n&&a[x][y]==0; } int di[4][2]={{0,1},{1,0},{0,-1},{-1,0}}; void solve2(){ int d=0; int sx=1,sy=1; memset(a,0,sizeof a); int cur=1; for(int it=1;it<=n*n;it++){ a[sx][sy]=cur++; int nx=sx+di[d][0],ny=sy+di[d][1]; if(it==n*n)break; while(!isok(nx,ny)){ d++; d%=4; nx=sx+di[d][0],ny=sy+di[d][1]; } sx=nx,sy=ny; } } int main(){ while(scanf("%d%d",&n,&ty)!=EOF){ if(ty==1||ty==2)solve1(); if(ty==3||ty==4)solve2(); if(ty==2||ty==4)rev(); for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ printf("%d%c",a[i][j],j==n?'\n':' '); } } } return 0; }
F. Buddy Numbers
若質數太多那麼顯然無解,對於小資料爆搜即可。
#include <bits/stdc++.h> using namespace std ; int a[1000] , n ; void solve () { if ( n == 1 ) { printf ( "1\n" ) ; return ; } if ( n == 2 ) { printf ( "1 2\n" ) ; return ; } if ( n == 3 ) { printf ( "2 1 3\n" ) ; return ; } if ( n == 4 ) { printf ( "2 4 1 3\n" ) ; return ; } if ( n == 6 ) { printf ( "3 6 2 4 1 5\n" ) ; return ; } printf ( "-1\n" ) ; } int main () { while ( ~scanf ( "%d" , &n ) ) solve () ; return 0 ; }
G. Gmoogle
按題意模擬即可。注意首尾空格的處理以及文末沒有標點符號的情況的處理。
#include<stdio.h> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<vector> #include<set> #include<map> #include<queue> #include<time.h> #include<assert.h> #include<sstream> #include<iostream> using namespace std; typedef long long LL; typedef pair<int,int>pi; string s; vector<string>jz; int can[1020]; int isend(int loc){ if(s[loc]!='.'&&s[loc]!='!'&&s[loc]!='?')return 0; if(s[loc]=='.'){ int j=loc+1; while(j<s.size()&&s[j]==' ')j++; if(j<s.size()&&s[j]>='a'&&s[j]<='z')return 0; } return 1; } void cg(char &c){if(c>='A'&&c<='Z')c=c-'A'+'a';} bool cmp(char c1,char c2){ cg(c1),cg(c2); if(c1==c2)return 1; return 0; } bool ok(string &wb,string& in){ for(int i=0;i+in.size()<=wb.size();i++)if(isalpha(wb[i])){ bool flag=1; if(i&&isalpha(wb[i-1]))continue;// int j=i+in.size(); if(j<wb.size()&&isalpha(wb[j]))continue;// //while ( j < wb.size () && wb[j] == ' ' ) ++ j ; //if(j<wb.size()-1&&wb[j]=='.')continue; for(int j=0;j<in.size();j++){ if(cmp(wb[i+j],in[j])!=1){flag=0;break;} } if(flag)return 1; } return 0; } bool zhu(char x){ if(x>='a'&&x<='z')return 1; if(x>='A'&&x<='Z')return 1; if(x==' ')return 1; if(x>='0'&&x<='9')return 1; if(x=='.'||x=='?'||x=='!')return 1; return 0; } void fix ( string& s ) { while ( !zhu(s[s.size () - 1]) ) s.pop_back () ; } int main(){ getline(cin,s); fix ( s ) ; int k = 0 ; for(int l=0;l<s.size();){ while(l<s.size()&&s[l]==' ')l++; if(l>=s.size())break; //if ( s[l] == '.' /*|| s[l] == '?' || s[k] == '!'*/ ) while ( 1 ) ; int r=l; string tmp=""; while(r<s.size()&&!isend(r))tmp.push_back(s[r++]); if ( r < s.size () ) tmp.push_back(s[r++]); while(tmp[tmp.size()-1]==' ')tmp.pop_back(); //printf ( "%d\n" , ( int ) tmp.size () ) ; jz.push_back(tmp); l=r; } /* cout<<"OUTPUT"<<endl; for(int i=0;i<jz.size();i++)cout<<jz[i]<<endl; */ int q;scanf("%d",&q); string rub; getline(cin,rub); while(q--){ string que; getline(cin,que); fix ( que ) ; for(int i=0;i<jz.size();i++)can[i]=1; stringstream ss(que); string tmp; while(ss>>tmp){ fix ( tmp ) ; for(int i=0;i<jz.size();i++){ if(!ok(jz[i],tmp))can[i]=0; } } int i=0,j=que.size()-1; while(que[i]==' ')i++; while(que[j]==' ')j--; cout<<"Search results for \""; for(int it=i;it<=j;it++)printf("%c",que[it]); cout<<"\":\n"; for(int i=0;i<jz.size();i++)if(can[i]){ cout<<"- \""<<jz[i]<<"\"\n"; } } return 0; }
H. Generator
線性篩求出每個數的最小質因子即可。
#include<stdio.h> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<vector> #include<set> #include<map> #include<queue> #include<time.h> #include<assert.h> #include<iostream> using namespace std; typedef long long LL; typedef pair<int,int>pi; LL n; LL sum[10000020]; int tot[10000020]; bool isp[10000200]; vector<int>pri; const int Maxn=10000010; int getsq(int x){ int tmp=sqrt(x+0.5); while(tmp*tmp<x)tmp++; while(tmp*tmp>x)tmp--; return tmp-1; } void pre(){ for(int i=2;i<Maxn;i++){ if(!isp[i])pri.push_back(i),sum[i]=getsq(i); for(int j=0;j<pri.size();j++){ int x=pri[j]; if(x*i>=Maxn)break; sum[x*i]=x-1; isp[i*x]=1; if(i%x==0)break; } } //for(int i=2;i<=10;i++)printf("%lld ",sum[i]);puts(""); for(int i=2;i<Maxn;i++)sum[i]+=sum[i-1]; for(int i=2;i<Maxn;i++)tot[i]=tot[i-1]+(!isp[i]); } int main(){ pre(); int _;scanf("%d",&_); while(_--){ scanf("%lld",&n); LL ans=sum[n]; if(ans==0)printf("0/1\n"); else{ LL gc=__gcd((LL)tot[n],ans); printf("%lld/%lld\n",ans/gc,tot[n]/gc); } } return 0; }
I. Addition
維護二進位制字典樹即可。
#include <bits/stdc++.h> using namespace std ; typedef long long LL ; const int MAXN = 100005 ; int nxt[MAXN][2] ; int cnt[MAXN] ; char s[MAXN] , s1[MAXN] ; int cur , root ; int n , m ; int newnode () { ++ cur ; nxt[cur][0] = nxt[cur][1] = 0 ; cnt[cur] = 0 ; return cur ; } void insert ( char s[] ) { int now = root ; for ( int i = 0 ; i < m ; ++ i ) { int x = s[i] ; if ( !nxt[now][x] ) nxt[now][x] = newnode () ; now = nxt[now][x] ; cnt[now] ++ ; } } int query ( char s[] ) { int now = root , ans = 0 ; for ( int i = 0 ; i < m ; ++ i ) { int x = s[i] ; if ( !x ) ans += cnt[nxt[now][1]] ; now = nxt[now][x] ; } return ans ; } void solve () { cur = 0 ; root = newnode () ; for ( int i = 1 ; i <= n ; ++ i ) { scanf ( "%s" , s ) ; for ( int j = 0 ; j < m ; ++ j ) { s[j] -= '0' ; s1[j] = 1 ^ s[j] ; } int ans = query ( s1 ) ; insert ( s ) ; printf ( "%d\n" , ans ) ; fflush ( stdout ) ; } } int main () { scanf ( "%d%d" , &n , &m ) ; solve () ; return 0 ; }
J. Votter and Paul De Mort
留坑。
K. GCD on the segments
考慮列舉左端點$i$,則隨著右端點的右移,一共只有$O(\log n)$種不同的$\gcd$取值。所以首先通過ST表+二分查詢預處理出$O(n\log n)$個四元組$(x,i,l,r)$,表示左端點為$i$,右端點取值範圍在$[l,r]$內,且這一段的$\gcd$都為$x$。
將四元組按照$x$為第一關鍵字,$i$為第二關鍵字排序,對於相同的$x$一起處理。
當$x$相同時,顯然所有的$i$互不相同。設$f[i]$為恰好以位置$i$為結尾的最優解,則對於一個四元組$(x,i,l,r)$,能更新它的最優解為區間$[1,i-1]$的最優值$+1$,然後用它更新區間$[l,r]$的$f[]$。用支援打標記的線段樹維護即可。時間複雜度$O(n\log^2n)$。
#include<cstdio> #include<cstdlib> #include<algorithm> using namespace std; const int N=100010,K=17,P=1000000007,M=262145; int T,n,m,i,j,x,y,l,r,mid,Log[N],val,f[K][N]; struct PI{ int x,i,l,r; PI(){} PI(int _x,int _i,int _l,int _r){x=_x,i=_i,l=_l,r=_r;} }a[3000000]; inline bool cmp(PI a,PI b){return a.x==b.x?a.i<b.i:a.x<b.x;} struct Num{ int x,y; Num(){x=y=0;} Num(int _x,int _y){x=_x,y=_y;} inline Num operator+(Num b){ if(x<b.x)return b; if(x>b.x)return Num(x,y); return Num(x,(y+b.y)%P); } inline Num operator+(int _x){return Num(x+_x,y);} inline Num operator-(int b){return Num(x,(long long)y*b%P);} inline void operator+=(Num b){*this=*this+b;} }tmp,v[M],tag[M],ans; int pos[M]; inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10LL)+=c-'0';} inline int askgcd(int y){int k=Log[y-i+1];return __gcd(f[k][i],f[k][y-(1<<k)+1]);} inline void clean(int x){ if(pos[x]<T)pos[x]=T,v[x]=tag[x]=Num(); } inline void tag1(int x,Num y){ clean(x); v[x]+=y; tag[x]+=y; } inline void pb(int x){ if(tag[x].x){ tag1(x<<1,tag[x]); tag1(x<<1|1,tag[x]); tag[x]=Num(); } } inline void up(int x){ clean(x<<1),clean(x<<1|1); v[x]=v[x<<1]+v[x<<1|1]; } void change(int x,int a,int b,int c,int d){ clean(x); if(c<=a&&b<=d){tag1(x,tmp);return;} pb(x); int mid=(a+b)>>1; if(c<=mid)change(x<<1,a,mid,c,d); if(d>mid)change(x<<1|1,mid+1,b,c,d); up(x); } void ask(int x,int a,int b,int c,int d){ clean(x); if(c<=a&&b<=d){tmp+=v[x];return;} pb(x); int mid=(a+b)>>1; if(c<=mid)ask(x<<1,a,mid,c,d); if(d>mid)ask(x<<1|1,mid+1,b,c,d); up(x); } int main(){ for(i=2;i<=100000;i++)Log[i]=Log[i>>1]+1; while(~scanf("%d",&n)){ m=0; ans=Num(); for(i=1;i<=n;i++)read(f[0][i]); int flag=0; for(i=2;i<=n;i++)if(f[0][i]!=f[0][i-1]){flag=1;break;} if(!flag){printf("%d 1\n",n);continue;} for(j=1;j<K;j++)for(i=1;i+(1<<j-1)<=n;i++)f[j][i]=__gcd(f[j-1][i],f[j-1][i+(1<<j-1)]); for(i=1;i<=n;i++)for(x=i;x<=n;x=y+1){ val=askgcd(y=x),l=x+1,r=n; while(l<=r)if(askgcd(mid=(l+r)>>1)==val)l=(y=mid)+1;else r=mid-1; a[++m]=PI(val,i,x,y); } sort(a+1,a+m+1,cmp); for(i=1;i<=m;i++){ if(i==1||a[i].x!=a[i-1].x)T++; tmp=Num(); if(a[i].i>1)ask(1,1,n,1,a[i].i-1); if(!tmp.x)tmp=Num(1,1);else tmp.x++; ans+=tmp-(a[i].r-a[i].l+1); change(1,1,n,a[i].l,a[i].r); } printf("%d %d\n",ans.x,ans.y); } return 0; }
L. Fibonacci Equation
分類討論。
#include<stdio.h> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<vector> #include<set> #include<map> #include<queue> #include<time.h> #include<assert.h> #include<iostream> using namespace std; typedef long long LL; typedef pair<int,int>pi; LL a[100]; LL x,y,z; int main(){ /* a[0]=0,a[1]=1; for(int i=2;i<=30;i++)a[i]=a[i-1]+a[i-2]; for(int i=1;i<=25;i++){ printf("%lld %lld\n",a[i],a[i]*a[i]-4*a[i-1]*a[i+1]); } */ while(scanf("%lld%lld%lld",&x,&y,&z)!=EOF){ if(x==0){printf("1\n");continue;} if(y>x&&y>z){ if(y==3)printf("1\n"); else printf("2\n"); } if(y<x&&y<z)printf("0\n"); if(y>min(x,z)&&y<max(x,z)){ if(y==1)printf("2\n"); else printf("0\n"); } //printf("real=%lld\n",a[y]*a[y]-4*a[x]*a[z]); } return 0; }