组队练习 2011年福州全国邀请赛

A 水题。

B 计算几何求重心，枚举对称点。

#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <cmath>
#include <iostream>
using namespace std;
#define maxn 600
#define eps 1e-6
bool zero(double a) {return (fabs(a)<eps?1:0);}
struct point{
    double x,y;
}po[maxn],center;
int n;
struct line{
    point a,b;
};
double cross1(point &a,point &b,point &c){
    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}
double cross(point p1,point p2,point p0){
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
point intersection(line u,line v){
    point ret=u.a;
    double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
    ret.x+=(u.b.x-u.a.x)*t;
    ret.y+=(u.b.y-u.a.y)*t;
    return ret;
}
point barycenter(point a,point b,point c){
    line u,v;
    u.a.x=(a.x+b.x)/2;
    u.a.y=(a.y+b.y)/2;
    u.b=c;
    v.a.x=(a.x+c.x)/2;
    v.a.y=(a.y+c.y)/2;
    v.b=b;
    return intersection(u,v);
}
point barycenter(int n,point *p){
    point ret,t;
    double t1=0,t2;
    int i;
    ret.x=ret.y=0;
    for(i=1;i<n-1;i++){
        if(fabs(t2=cross(p[0],p[i],p[i+1]))>eps){
            t=barycenter(p[0],p[i],p[i+1]);
            ret.x+=t.x*t2;
            ret.y+=t.y*t2;
            t1+=t2;
        }
    }
    if(fabs(t1)>eps)
        ret.x/=t1,ret.y/=t1;
    return ret;
}
double dmult(point &a,point &b){
    return a.x*b.x+a.y*b.y;
}
double dmult(double a,double b,double c,double d){
    return a*c+b*d;
}
bool is_ok(int l,int r,int a,int b){
    if(r<l)return true;
    point temp,t;
    while(l<r){
        temp.x=(po[l].x+po[r].x)/2;
        temp.y=(po[l].y+po[r].y)/2;
         if(!zero(dmult(po[l].x-po[r].x,po[l].y-po[r].y,temp.x-center.x,temp.y-center.y)))
            return false;
         l++,r--;
    }
    temp.x=(po[a].x+po[b].x)/2;
    temp.y=(po[a].y+po[b].y)/2;
    if(l==r)
    return zero(cross(temp,center,po[l]));
    return true;
}
bool sovle(){
    int i,k;
    point temp;
    for(i=1;i<n;i++){
        temp.x=(po[0].x+po[i].x)/2;
        temp.y=(po[0].y+po[i].y)/2;
        if(zero(dmult(po[i].x-po[0].x,po[i].y-po[0].y,temp.x-center.x,temp.y-center.y))){
                if(is_ok(1,i-1,0,i)&& is_ok(i+1,n-1,0,i))
                return true;
        }
    }
     for(i=2;i<n;i++){
        temp.x=(po[1].x+po[i].x)/2;
        temp.y=(po[1].y+po[i].y)/2;
        if(zero(dmult(po[i].x-po[1].x,po[i].y-po[1].y,temp.x-center.x,temp.y-center.y))){
            //printf("HHHH%lf %lf
",po[i].x,po[i].y);
                if(is_ok(2,i-1,1,i)&& is_ok(i+1,n,1,i))
                return true;
        }
    }
    return false;
}
int main(){
    int i,j,k,t,cas=0;
    scanf("%d",&t);
    while(t--){

        scanf("%d",&n);
        for(i=0;i<n;i++)
        scanf("%lf%lf",&po[i].x,&po[i].y);
        k=1;
        for(i=2;i<n;i++){
            if(zero(cross(po[k-1],po[k],po[i])))
            po[k]=po[i];
            else po[++k]=po[i];
        }
        k++;
        if(zero(cross(po[0],po[k-1],po[k-2])))
        k--;
        if(zero(cross(po[k-1],po[0],po[1]))){
                for(i=1;i<k;i++)
                 po[i-1]=po[i];
                 k--;
        }
        n=k;
        printf("Case %d: ",++cas);
        if(k<=2){
            printf("YES
");
            continue;
        }
        po[n]=po[0];
        center=barycenter(n,po);
        if(sovle())
            printf("YES
");
        else
            printf("NO
");
    }
    return 0;
}

C 数学题。

import java.math.*;
import java.math.BigDecimal;
import java.util.*;
public class Main {
    public static void main(String arg[]){
        BigDecimal a,b,c,x,y,z,d,an,ans;
        Scanner cin=new Scanner(System.in);
        int t=1,tt;
        tt=cin.nextInt();
        an=BigDecimal.valueOf(32);
        while(tt-->0){
            a=cin.nextBigDecimal();
            b=cin.nextBigDecimal();
            if(a.compareTo(b)>0){
                c=a.subtract(b);
                d=b;
                ans=a;
            }else{
                c=b.subtract(a);
                d=a;
                ans=b;
            }
            if(c.compareTo(an)<0){
                double s=c.doubleValue();
                s=Math.pow(2.0,s)+1;
                s=Math.log(s)/Math.log(2.0);
                ans=d;
                ans=ans.add(BigDecimal.valueOf(s));
            }
            ans = ans.setScale(9, BigDecimal.ROUND_HALF_UP);
            System.out.println("Case "+t+": "+ans);
            t++;
        }
    }
}

D 数学

#include<iostream>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<iomanip>
#include<cstdio>
#define ll long long
#define M 1000001
#define mod 1000000007
using namespace std;
ll an[M];
double p[M];
void init()
{
    an[0]=0;an[1]=1;
    p[0]=0;p[1]=1;
    ll t=1;
    for(int i=2;i<M;i++){
        an[i]=(i*an[i-1]%mod+t)%mod;
        t=(t*i)%mod;
        p[i]=p[i-1]+1.0/(double)i;
    }
}
int main()
{
    int t,n,ca=0;
    init();
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        printf("Case %d: %I64d %.6lf
",++ca,an[n],p[n]);
    }
    return 0;
}

E 树形DP。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define Maxn 100010
#define LL __int64
using namespace std;
int son[Maxn],vi[Maxn],head[Maxn],e;
struct Edge{
    int u,v,next;
    LL val;
}edge[Maxn*3];
LL ans,sum;
void add(int u,int v,LL val)
{
    edge[e].u=u,edge[e].v=v,edge[e].val=val,edge[e].next=head[u],head[u]=e++;
    edge[e].u=v,edge[e].v=u,edge[e].val=val,edge[e].next=head[v],head[v]=e++;
}
void init()
{
    memset(son,0,sizeof(son));
    memset(vi,0,sizeof(vi));
    memset(head,-1,sizeof(head));
    e=0;
}
void dfs(int u)
{
    int i,v;
    son[u]=1;
    vi[u]=1;
    for(i=head[u];i!=-1;i=edge[i].next){
        v=edge[i].v;
        //cout<<u<<" "<<v<<" ";
        if(vi[v]) continue;
        dfs(v);
        //cout<<son[v]<<" "<<sum<<" "<<edge[i].val<<endl;
        ans+=son[v]*(sum-son[v])*2*edge[i].val;
        son[u]+=son[v];
    }
}
int main()
{
    int t,n,a,b,i,j,Ca=0;
    LL c;
    scanf("%d",&t);
    while(t--){
        init();
        scanf("%d",&n);
        for(i=1;i<n;i++){
            scanf("%d%d%I64d",&a,&b,&c);
            a++,b++;
            add(a,b,c);
        }
        ans=0;
        sum=n;
        dfs(1);
        printf("Case %d: %I64d
",++Ca,ans);
    }
    return 0;
}

F 是个最大匹配

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#define Maxn 110
using namespace std;
int match[Maxn],n,m;
bool g[Maxn][Maxn],vi[Maxn];
int dfs(int u)
{
    int i;
    for(i=1;i<=m;i++){
        if(!vi[i]&&g[u][i]){
            vi[i]=1;
            if(match[i]==-1||dfs(match[i])){
                match[i]=u;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    int t,e,i,j,u,v,Ca=0;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d%d",&n,&m,&e);
        memset(g,1,sizeof(g));
        for(i=1;i<=e;i++){
            scanf("%d%d",&u,&v);
            g[u][v]=0;
        }
        memset(match,-1,sizeof(match));
        int ans=0;
        for(i=1;i<=n;i++){
            memset(vi,0,sizeof(vi));
            if(dfs(i))
                ans++;
        }
        printf("Case %d: %d
",++Ca,ans);
    }
    return 0;
}

 先O（n）预处理出ri[i][j],le[i][j]，分别表示第i个位置向右边移动出j个空格需要的步数，表示第i个位置向左边移动出j个空格需要的步数。

然后枚举间隙处，二分判段最大间隔。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define Maxn 710
#define inf 100000000
using namespace std;
int ri[Maxn][Maxn],le[Maxn][Maxn],n,m;
char str[Maxn];
bool OK(int pos,int x)
{
    int i,j,cnt=inf;
    if(pos==1) {
        cnt=min(cnt,ri[pos][x]);
        return cnt<=m;
    }
    for(i=0;i<=n;i++){
        if(i>x) break;
        cnt=min(cnt,ri[pos][i]+le[pos-1][x-i]);
    }
    //cout<<pos<<" "<<x<<" "<<cnt<<" "<<m<<endl;
    return cnt<=m;
}
int main()
{
    int t,i,j,ze,Ca=0;
    scanf("%d",&t);
    while(t--){
        for(i=1;i<Maxn;i++){
            for(j=1;j<Maxn;j++){
                ri[i][j]=le[i][j]=inf;
            }
        }
        ze=0;
        scanf("%d%d%s",&n,&m,str);
        if(str[0]=='0') le[1][1]=0,ze=1;
        for(i=1;i<n;i++){
            if(str[i]=='0') ze++;
            for(j=1;j<=n;j++){
                if(le[i][j-1]>=inf) break;
                if(str[i]=='1')
                    le[i+1][j]=le[i][j]+j;
                else
                    le[i+1][j]=le[i][j-1];
            }
        }
        if(str[n-1]=='0') ri[n][1]=ri[n+1][1]=0;
        for(i=n-1;i>=1;i--){
            for(j=1;j<=n;j++){
                if(ri[i+1][j-1]>=inf) break;
                if(str[i-1]=='1')
                    ri[i][j]=ri[i+1][j]+j;
                else
                    ri[i][j]=ri[i+1][j-1];
            }
        }
        int f=0;
        int ans=0,l,r,mid;
        for(i=1;i<=n;i++){
            if(str[i-1]=='0'){
                l=0;r=ze;
                while(l+1<r){
                    mid=(l+r)>>1;
                    if(OK(i,mid))
                        l=mid;
                    else
                        r=mid;
                }
                ans=max(ans,l);
                if(OK(i,r))
                    ans=max(ans,r);
            }
        }
        printf("Case %d: %d
",++Ca,ans);
    }
    return 0;
}