2015亚洲区北京站网络赛

A http://hihocoder.com/problemset/problem/1227

题意：给出平面上m个点，任选一个点为圆心，选最小的半径使得恰好n个点在圆内。

解法：枚举圆心，然后对所有点到该圆心距离排序，取第n大的距离作为半径，若第n+1大的在圆内或圆上不合法。

 

//#define debug
//#define txtout
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define mt(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double eps=1e-8;
const double pi=acos(-1.0);
const int inf=0x3f3f3f3f;
const int M=1e2+10;
struct point {
    double x,y;
} p[M];
double dist[M];
int n,m;

double Square(double x) { ///平方
    return x*x;
}
double Distance(point a,point b) { ///平面两点距离
    return sqrt(Square(a.x-b.x)+Square(a.y-b.y));
}

int choose(point c) {
    for(int i=0; i<m; i++) {
        dist[i]=Distance(c,p[i]);
    }
    sort(dist,dist+m);
    double r=dist[n-1];
    int int_r=(int)(r+1);
    if(n<m&&dist[n]<int_r+eps)
        return inf;
    return int_r;
}
int solve() {
    if(n>m) return -1;
    int res=inf;
    for(int i=0; i<m; i++) {
        res=min(res,choose(p[i]));
    }
    if(res==inf) res=-1;
    return res;
}
int main() {
#ifdef txtout
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif
    int t;
    while(~scanf("%d",&t)) {
        while(t--) {
            scanf("%d%d",&m,&n);
            for(int i=0; i<m; i++) {
                scanf("%lf%lf",&p[i].x,&p[i].y);
            }
            printf("%d\n",solve());
        }
    }
    return 0;
}

 

B http://hihocoder.com/problemset/problem/1228

题意：模拟一行文本的输入操作。输入文本长度限制m，操作序列string，输出最后的文本。

操作L将光标左移，R光标右移，S切换插入和复写的输入模式，D删右边一个或选中区间，B删左边一个，C选区间，V将缓冲区复制一份。完全符合计算机复制粘贴等，细心模拟题。 

解法：读题，模拟。

 

//#define debug
//#define txtout
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define mt(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double eps=1e-8;
const double pi=acos(-1.0);
const int inf=0x3f3f3f3f;
const int M=1e4+10;
char input[M];
char output[M];
char buffer[M];
char save[M];
int limit;
int length_of_output;
int length_of_buffer;
int index_of_caret;
int index_of_copy;
bool overwrite_mode;
bool start;
void init(){
    index_of_caret=0;
    length_of_output=0;
    overwrite_mode=false;
    start=false;
}
void Insert(int id,char c){
    length_of_output++;
    for(int i=length_of_output-1;i>=id;i--){
        output[i]=output[i-1];
    }
    output[id]=c;
}
void Delete(int x,int y){
    for(int i=y+1;i<length_of_output;i++){
        output[x+i-y-1]=output[i];
    }
    length_of_output-=(y-x+1);
}
void Copy(int x,int y){
    length_of_buffer=0;
    for(int i=x;i<=y;i++){
        buffer[length_of_buffer++]=output[i];
    }
}
void Flip(bool &a){
    a=!a;
}
void solve_L(){
    if(index_of_caret) index_of_caret--;
}
void solve_R(){
    if(index_of_caret<length_of_output) index_of_caret++;
}
void solve_S(){
    Flip(overwrite_mode);
}
void solve_D(){
    if(start){
        if(index_of_caret<index_of_copy){
            Delete(index_of_caret,index_of_copy-1);
        }
        else{
            Delete(index_of_copy,index_of_caret-1);
            index_of_caret=index_of_copy;
        }
        Flip(start);
        return ;
    }
    if(index_of_caret<length_of_output){
        Delete(index_of_caret,index_of_caret);
    }
}
void solve_B(){
    if(index_of_caret){
        Delete(index_of_caret-1,index_of_caret-1);
        index_of_caret--;
    }
}
void solve_C(){
    if(start){
        if(index_of_caret<index_of_copy){
            Copy(index_of_caret,index_of_copy-1);
        }
        else{
            Copy(index_of_copy,index_of_caret-1);
        }
    }
    else{
        index_of_copy=index_of_caret;
        length_of_buffer=0;
    }
    Flip(start);
}
void solve_V(){
    if(overwrite_mode){
        if(index_of_caret+length_of_buffer>limit) return ;
        for(int i=0;i<length_of_buffer;i++){
            output[index_of_caret+i]=buffer[i];
        }
        index_of_caret+=length_of_buffer;
        length_of_output=max(length_of_output,index_of_caret);
    }
    else{
        if(length_of_output+length_of_buffer>limit) return ;
        int length_of_save=0;
        for(int i=index_of_caret;i<length_of_output;i++){
            save[length_of_save++]=output[i];
        }
        for(int i=0;i<length_of_buffer;i++){
            output[index_of_caret+i]=buffer[i];
        }
        index_of_caret+=length_of_buffer;
        for(int i=0;i<length_of_save;i++){
            output[index_of_caret+i]=save[i];
        }
        length_of_output+=length_of_buffer;
    }
}
void add(char c){
    if(overwrite_mode){
        if(index_of_caret==length_of_output&&length_of_output==limit) return ;
        output[index_of_caret]=c;
        if(index_of_caret==length_of_output) length_of_output++;
    }
    else{
        if(length_of_output==limit) return ;
        Insert(index_of_caret,c);
    }
    index_of_caret++;
}
void solve_by_type(char c){
    if(c=='L') return solve_L();
    if(c=='R') return solve_R();
    if(c=='D') return solve_D();
    if(c=='C') return solve_C();
    start=false;
    if(c=='B') return solve_B();
    if(c=='S') return solve_S();
    if(c=='V') return solve_V();
    return add(c);
}
void solve(){
    init();
    for(int i=0;input[i];i++){
        solve_by_type(input[i]);
    }
    output[length_of_output]=0;
    if(length_of_output==0) strcpy(output,"NOTHING");
}
int main(){
    #ifdef txtout
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif
    int t;
    while(~scanf("%d",&t)){
        while(t--){
            scanf("%d%s",&limit,input);
            solve();
            puts(output);
        }
    }
    return 0;
}

 

 H http://hihocoder.com/problemset/problem/1234

题意：给一个边长1的正方形，每次选每条边中点构成新的正方形，重复1000次，输入一个坐标k，求x=k这条直线与该图形交点个数。

解法：精度至少1e-10，枚举前30个竖线的x坐标存数组，然后测输入的坐标在哪个范围，每往前走一个范围答案+4，如果离0.5非常近，就是2000.

 

//#define debug
//#define txtout
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<ctime>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define mt(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long LL;
const double eps=1e-10;
const double pi=acos(-1.0);
const int inf=0x3f3f3f3f;
const int M=1e2+10;
double x;
double y[M];
bool zero(double x){
    return (x>0?x:-x)<eps;
}
void init(){
    double add=0.25;
    y[0]=0;
    for(int i=1;i<30;i++){
        y[i]=y[i-1]+add;
        add*=0.5;
    }
}
int solve(){
    int res=0;
    for(int i=0;i<30;i++){
        if(zero(x-y[i])) return -1;
        if(x<y[i]) return res;
        res+=4;
    }
    return 2000;
}
int main(){
    #ifdef txtout
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif
    init();
    int t;
    while(~scanf("%d",&t)){
        while(t--){
            scanf("%lf",&x);
            printf("%d\n",solve());
        }
    }
    return 0;
}

 

 

end