AtCoder Beginner Contest 151 题解

竟然顺利地AK了，爽到（
传送门：https://atcoder.jp/contests/abc151

A

语法题

#include<bits/stdc++.h>
using namespace std;

int main(){
	char ch; cin>>ch;
	cout<<(char)(ch+1);
	return 0;
}

B

简单的判断

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

int main(){
	int n, k, m; cin>>n>>k>>m;
	
	int sum=0;
	rep(i,1,n-1){
		int t; cin>>t;
		sum+=t;
	}
	
	if(sum+k<n*m) puts("-1");
	else{
		cout<<(n*m-sum>0? n*m-sum: 0)<<endl;
	}
    return 0;
}

C

简单的模拟

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=1e5+5;
bool ok[N];

int main(){
	int n, m; cin>>n>>m;
	
	int ac=0, pe[n+1]={0};
	rep(i,1,m){
		int id; string op; cin>>id>>op;
		if(op=="AC" && !ok[id]){
			ok[id]=true;
			ac++;
		}
		else if(op=="WA" && !ok[id]) pe[id]++;
	}
	
	int pen=0;
	rep(i,1,n) if(ok[i]) pen+=pe[i];
	cout<<ac<<' '<<pen<<endl;
	
    return 0;
}

D

看到范围，直接上bfs即可

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

#define x first
#define y second

const int N=25;

int n, m; 
char g[N][N];

PII q[N*N];
bool vis[N][N];
int d[N][N];

int bfs(int sx, int sy, int tx, int ty){
	if(sx==tx && sy==ty) return 0;
	memset(vis, false, sizeof vis);
	memset(d, 0x3f, sizeof d);
	int hh=0, tt=-1;
	q[++tt]={sx, sy}, d[sx][sy]=0;
	
	int dx[]={1, 0, -1, 0}, dy[]={0, 1, 0, -1};
	
	while(tt>=hh){
		auto hd=q[hh++];
		int x=hd.x, y=hd.y;
		rep(i,0,3){
			int kx=x+dx[i], ky=y+dy[i];
			if(g[kx][ky]=='#' || vis[kx][ky]) continue;
			if(kx<1 || kx>n || ky<1 || ky>m) continue;
			
			vis[kx][ky]=true;
			d[kx][ky]=d[x][y]+1;
			if(kx==tx && ky==ty) return d[kx][ky];
			q[++tt]={kx, ky};
		}
	}
	return -1;
}

int main(){
	cin>>n>>m;
	rep(i,1,n) rep(j,1,m) cin>>g[i][j];
	
	int res=-1;
	rep(i,1,n) rep(j,1,m) rep(r,1,n) rep(c,1,m){
		if(g[i][j]=='#' || g[r][c]=='#') continue;
		res=max(res, bfs(i, j, r, c));
	}
	cout<<res<<endl;
	
    return 0;
}

E

考虑每个数可以造成的贡献，我们考察第 (i) 个数（这个数本身自然是被选取了），有三种情况：

• 如果这个数的左右的数都有选取，那么贡献为 (0)
• 如果选取的数在该数左边，那么贡献为 (C_{i-1}^{k-1}w_i)
• 如果选取的数在该数右边，那么贡献为 (-C_{n-i}^{k-1}w_i)

直接统计就行了：

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=1e5+5, mod=1e9+7;

ll fpow(ll x,ll p)
{
    ll res=1;
    for(;p;p>>=1,x=x*x%mod)
        if(p&1)res=res*x%mod;
    return res%mod;
}

ll inv(ll x){
	return fpow(x,mod-2)%mod;
}

ll fac[N];

void init(){
	fac[0]=1;
	for(int i=1; i<N; i++) fac[i]=fac[i-1]*i%mod;
}

ll C(ll a, ll b){
	return fac[a]*inv(fac[b])%mod*inv(fac[a-b])%mod;
}

int w[N];

int main(){
	init();
	int n, k; cin>>n>>k;
	rep(i,1,n) cin>>w[i];
	
	sort(w+1, w+1+n);
	
	ll res=0;
	rep(i,1,n){
		ll a=0, b=0;
		if(k-1<=i-1) a=C(i-1, k-1);
		if(n-i>=k-1) b=C(n-i, k-1);
		res=(1LL*(a-b)*w[i]+res)%mod;
	}
	cout<<res<<endl;
	
    return 0;
}

F

似乎是最小圆覆盖的板子题，但我直接用模拟退火搞了hh

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

#define x first
#define y second

const double eps=1e-12;

const int N=55;
int n;
PDD q[N];

double ans=1e5;

double rand(double l, double r){
    return (double)rand()/RAND_MAX*(r-l)+l;
}

double get_dist(PDD u, PDD v){
	return sqrt((u.x-v.x)*(u.x-v.x)+(u.y-v.y)*(u.y-v.y));
}

double calc(PDD p){
	double res=0;
	rep(i,1,n) res=max(res, get_dist(p, q[i]));
	ans=min(ans, res);
	return res;
}

void anneal(){
	double ax=0, ay=0;
	rep(i,1,n) ax+=q[i].x, ay+=q[i].y;
	ax/=n, ay/=n;
	
	PDD cur(ax, ay);
	for(double t=500; t>1e-4; t*=0.992){
		PDD np(rand(cur.x-t, cur.x+t), rand(cur.y-t, cur.y+t));
		double dt=calc(np)-calc(cur);
		if(exp(-dt/t)>rand(0, 1)) cur=np;
	}
}

int main(){
	cin>>n;
	rep(i,1,n){
		double x, y; cin>>x>>y;
		x-=500, y-=500;
		q[i]={x, y};
	}
	random_shuffle(q+1, q+1+n);
	
	rep(i,1,100) anneal();
	printf("%.10lf", ans);
	
    return 0;
}

当然，因为就是一个最裸的最小圆覆盖问题，直接用最小圆覆盖板子就能过了：

#include<bits/stdc++.h>
using namespace std;

const double eps=1e-12, pi=acos(-1);

/*start-----------------------------------------------------------------*/

#define x first
#define y second

struct Point{
	double x, y;
	Point(double x=0, double y=0): x(x), y(y){}
	
	Point operator + (const Point &p)const {return Point(x+p.x, y+p.y);}
	Point operator - (const Point &p)const {return Point(x-p.x, y-p.y);}
	Point operator * (const Point &p)const {return Point(x*p.x, y*p.y);}
	Point operator / (const Point &p)const {return Point(x/p.x, y/p.y);}
	
	Point operator * (const double &k){return Point(x*k, y*k);}
	Point operator / (const double &k){return Point(x/k, y/k);}	
};

struct Circle{
	Point p;
	double r;
	Circle(Point p={0, 0}, double r=0): p(p), r(r){}
};

typedef Point Vector;

int sign(double x){
	if(fabs(x)<eps) return 0;
	return x<0? -1: 1;
}

int cmp(double x, double y){
	return sign(x-y);
}

bool operator < (const Point &a, const Point &b) {
	return sign(a.x-b.x)<0 || sign(sign(a.x-b.x)==0 && sign(a.y-b.y)<0);
}

double dot(Vector A, Vector B){
	return A.x*B.x+A.y*B.y;
}

double cross(Vector A, Vector B){
	return A.x*B.y-A.y*B.x;
}

double get_length(Vector A){
	return sqrt(dot(A, A));
}

double get_dist(Point a, Point b){
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double get_angle(Vector A, Vector B){
	return acos(dot(A, B)/get_length(A)/get_length(B));
}

double area(Point a, Point b, Point c){
	return cross(b-a, c-a);
}

// 向量 A 顺时针旋转 angle 度
Vector rotate(Vector A, double angle){
	return Vector(A.x*cos(angle)+A.y*sin(angle), -A.x*sin(angle)+A.y*cos(angle));
}

Point get_line_intersection(Point p, Vector v, Point q, Vector w){
	Vector u=p-q;
	double t=cross(w, u)/cross(v, w);
	return p+v*t;
}

// get perpendicular bisector 得到中垂线
pair<Point, Vector> get_pbline(Point a, Point b){
	return {(a+b)/2, rotate(b-a, pi/2)};
}

// 以两点为对径点作圆
Circle get_circle(Point a, Point b){
	return {(a+b)/2, get_dist(a, b)/2};
}

// 过三点作圆
Circle get_circle(Point a, Point b, Point c){
	auto u=get_pbline(a, b), v=get_pbline(a, c);
	Point p=get_line_intersection(u.x, u.y, v.x, v.y);
	return {p, get_dist(p, a)};
}

/*end--------------------------------------------*/

const int N=55;

Point q[N];
int n;

int main(){
	cin>>n;
	for(int i=1; i<=n; i++){
		double x, y; cin>>x>>y;
		q[i]={x, y};
	}
	
	random_shuffle(q+1, q+1+n);
	
	Circle C={q[1], 0};
	for(int i=2; i<=n; i++) if(cmp(C.r, get_dist(C.p, q[i]))==-1){
		C={q[i], 0};
		for(int j=1; j<i; j++) if(cmp(C.r, get_dist(C.p, q[j]))==-1){
			C=get_circle(q[i], q[j]);
			for(int k=1; k<j; k++) if(cmp(C.r, get_dist(C.p, q[k]))==-1){
				C=get_circle(q[i], q[j], q[k]);
			}
		}
	}
	
	printf("%.10lf
", C.r);
	return 0;
}