【ABC】151題解

阿新 • • 發佈：2021-06-15

竟然順利地AK了，爽到（

A

語法題

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

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

B

簡單的判斷

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#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

簡單的模擬

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#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

看到範圍，直接上bfs即可

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#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

考慮每個數可以造成的貢獻，我們考察第 $i$ 個數（這個數本身自然是被選取了），有三種情況：

如果這個數的左右的數都有選取，那麼貢獻為 $0$
如果選取的數在該數左邊，那麼貢獻為 $C_{i-1}^{k-1}w_i$
如果選取的數在該數右邊，那麼貢獻為 $-C_{n-i}^{k-1}w_i$

直接統計就行了：

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#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

似乎是最小圓覆蓋的板子題，但我直接用模擬退火搞了hh

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#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;
}