La Salle-Pui Ching Programming Challenge 培正喇沙編程挑戰賽 2017

A Ambiguous Dates

贪心，从小到大取日期

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (123456)ll d[MAXN];ll Abs(ll x){return max(x,-x);}int main(){//  freopen("A.in","r",stdin);//  freopen(".out","w",stdout);    int n=read();    For(i,n) d[i]=read();    ll ans=0;    sort(d+1,d+1+n);ll c=0;    For(i,n) {        ll p=d[i];        if (p>i) {            ans+=min((ll)n-i,d[i]-i);        }     }    cout<<2*ans<<endl;    return 0;}

B Bacteria Experiment

求树直径

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (912345)vi e[MAXN];int dep[MAXN]={};void dfs(int x,int fa) {    dep[x]=dep[fa]+1;    for(auto v:e[x]) {        if (v!=fa) dfs(v,x);    }}int main(){//  freopen("B.in","r",stdin);//  freopen(".out","w",stdout);    int n=read();    For(i,n-1) {        int x=read(),y=read();        e[x].pb(y);e[y].pb(x);    }    dfs(1,0);    int mx=max_element(dep+1,dep+1+n)-dep;    MEM(dep);    dfs(mx,0);    int l=*max_element(dep+1,dep+1+n);    int ans=0,dis=1; --l;    while(dis<l) ans++,dis<<=1;    cout<<ans<<endl;    return 0;}

C Cheering

签到

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} char s[12345];int main(){//  freopen("E.in","r",stdin);//  freopen(".out","w",stdout);    cin>>s;    int n=strlen(s),p=0;    Rep(i,n-2) {        if (s[i]=='L'&&s[i+1]=='S'&&s[i+2]=='C') ++p;    }    Rep(i,n-3) {        if (s[i]=='P'&&s[i+1]=='C'&&s[i+2]=='M'&&s[i+3]=='S') --p;    }    if (!p) puts("Tie");else puts(p>0?"LSC":"PCMS");    return 0;}

H Hit!

给2个圆保证存在不为0的公共部分，求公共部分的任意一个点
分2种情况，内含，内切（此时解取1个圆的圆心），相交圆心连线上的一个点。

#include<bits/stdc++.h>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=Pre[x];p;p=Next[p])#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define pb push_back#define mp make_pair #define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define ALL(x) (x).begin(),(x).end()typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} ll sqr(ll a){return a*a;}ld sqr(ld a){return a*a;}double sqr(double a){return a*a;}const double eps=1e-10;int dcmp(double x) {    if (fabs(x)<eps) return 0; else return x<0 ? -1 : 1; }ld PI = 3.141592653589793238462643383;class P{public:    double x,y;    P(double x=0,double y=0):x(x),y(y){}    friend ld dis2(P A,P B){return sqr(A.x-B.x)+sqr(A.y-B.y);   }    friend ld Dot(P A,P B) {return A.x*B.x+A.y*B.y; }    friend ld Length(P A) {return sqrt(Dot(A,A)); }    friend ld Angle(P A,P B) {        if (dcmp(Dot(A,A))==0||dcmp(Dot(B,B))==0||dcmp(Dot(A-B,A-B))==0) return 0;        return acos(max((ld)-1.0, min((ld)1.0, Dot(A,B) / Length(A) / Length(B) )) );     }    friend P operator- (P A,P B) { return P(A.x-B.x,A.y-B.y); }    friend P operator+ (P A,P B) { return P(A.x+B.x,A.y+B.y); }    friend P operator* (P A,double p) { return P(A.x*p,A.y*p); }    friend P operator/ (P A,double p) { return P(A.x/p,A.y/p); }    friend bool operator< (const P& a,const P& b) {return dcmp(a.x-b.x)<0 ||(dcmp(a.x-b.x)==0&& dcmp(a.y-b.y)<0 );}}; P read_point() {    P a;    scanf("%lf%lf",&a.x,&a.y);    return a;   } bool operator==(const P& a,const P& b) {    return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y) == 0;} typedef P V;double Cross(V A,V B) {return A.x*B.y - A.y*B.x;}double DistanceToLine(P p,P A,P B) {    V v1 = B-A, v2 = p-A;    return fabs(Cross(v1,v2))/Length(v1);}P GetLineProjection(P p,P A,P B) {    V v=B-A;    return A+v*(Dot(v,p-A)/Dot(v,v));}struct C{    P c;    double r,x,y;    C(){}    C(P c,double r):c(c),r(r),x(c.x),y(c.y){}    P point(double a) {        return P(c.x+cos(a)*r,c.y+sin(a)*r);    }};struct Line{    P p;    V v;    double ang;    Line(){}    Line(P p,V v):p(p),v(v) {ang=atan2(v.y,v.x); }    bool operator<(const Line & L) const {        return ang<L.ang;    }    P point(double a) {        return p+v*a;    }};int getLineCircleIntersection(Line L,C cir,double &t1,double &t2,vector<P> & sol) {    if (dcmp(DistanceToLine(cir.c,L.p,L.p+L.v)-cir.r)==0) {        P A=GetLineProjection(cir.c,L.p,L.p+L.v);        sol.pb(A);         t1 = (A-L.p).x / L.v.x;          return 1;    }    double a = L.v.x, b = L.p.x - cir.c.x, c = L.v.y, d= L.p.y - cir.c.y;    double e = a*a+c*c, f = 2*(a*b + c*d), g = b*b+d*d-cir.r*cir.r;    double delta = f*f - 4*e*g;    if (dcmp(delta)<0) return 0;    else if (dcmp(delta)==0) {        t1 = t2 = -f / (2*e); sol.pb(L.point(t1));        return 1;    }     t1 = (-f - sqrt(delta)) / (2*e); sol.pb(L.point(t1));    t2 = (-f + sqrt(delta)) / (2*e); sol.pb(L.point(t2));    return 2;}double angle(V v) {return atan2(v.y,v.x);}int getCircleCircleIntersection(C C1,C C2,vector<P>& sol) {    double d = Length(C1.c-C2.c);    if (dcmp(d)==0) {        if (dcmp(C1.r - C2.r)==0) return -1; //2圆重合         return 0;    }    if (dcmp(C1.r+C2.r-d)<0) return 0;    if (dcmp(fabs(C1.r-C2.r)-d)>0) return 0;    double a = angle(C2.c-C1.c);    double da = acos((C1.r*C1.r+d*d - C2.r*C2.r)/ (2*C1.r*d));    P p1 = C1.point(a-da), p2 = C1.point(a+da);    sol.pb(p1);    if (p1==p2) return 1;    sol.pb(p2);    return 2; }int main(){//  freopen("H.in","r",stdin);//  freopen(".out","w",stdout);    double x,y,r,a,b,c;    cin>>x>>y>>r>>a>>b>>c;    P p1=P(x,y),p2=P(a,b);    C c0=C(p1,r),c1=C(p2,c);    vector<P> sol;    if (getCircleCircleIntersection(c0,c1,sol)>=2) {        printf("%.10lf %.10lf\n",(sol[0].x+sol[1].x)/2 ,(sol[0].y+sol[1].y)/2 );     }    else if (r<c) {        printf("%.10lf %.10lf\n",x,y );     }    else {        printf("%.10lf %.10lf\n",a,b );     }    return 0;}

Inverted Signs

显然翻转一段只会影响2端点的贡献

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (1123456)ll h[MAXN],s[MAXN];ll Abs(ll x){return max(x,-x);}int main(){//  freopen("I.in","r",stdin);//  freopen(".out","w",stdout);    int n=read();    For(i,n) h[i]=read();    ll ans=0;    For(i,n-1) ans+=Abs(h[i]-h[i+1]);    s[1]=0;    Fork(i,2,n) {        s[i]=Abs(-h[i]-h[i-1]) - Abs(h[i]-h[i-1]);        s[i]=min(s[i],s[i-1]);    }    ll del=0,pcost=0;    ForD(i,n) {        ll p;        if (i==n) p=0;        else p=Abs(-h[i]-h[i+1]) - Abs(h[i]-h[i+1]);        pcost=min(p,pcost);        del= min (del , pcost+s[i]);    }    cout<<ans+del<<endl;    return 0;}

K Knights

四个角无法在第二阶段被其他Knights影响，只能自己覆盖自己。
覆盖了4个角以后能覆盖原图。

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} #define MAXN (123456)int b[2][2]={};int main(){//  freopen("K.in","r",stdin);//  freopen(".out","w",stdout);    int n=read(),m=read(),k=read();    For(i,k) {        int x=read(),y=read();        if ((1<x && x<n) || (1<y&&y<m) ) continue;        if(1==x&&1==y) b[0][0]=1;        if(1==x&&m==y) b[0][1]=1;        if(n==x&&1==y) b[1][0]=1;        if(n==x&&m==y) b[1][1]=1;    }    if (n==1 &&m==1) {        puts(b[1][1]?"0":"1");    }    else if (n==1) {        cout<<2-b[0][0]-b[0][1]<<endl;    }    else if (m==1) {        cout<<2-b[0][0]-b[1][0]<<endl;    }else {        cout<<4-b[0][0]-b[1][0]-b[0][1]-b[1][1]<<endl;    }    return 0;}