AtCoder ABC142

本来不想发这篇博客的
但是考虑到这是人生中第一次atcoder abc做出5道题
还是忍不住发一波

T1

看到了就很慌
作为史上做过的最难得abcT1，居然不是像140那种的直接输出$N^3$
翻译一下题意，输出$\frac{小于n的奇数个数}{n}$

# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
using namespace std;

# define Rep(i,a,b) for(int i=a;i<=b;i++)
# define _Rep(i,a,b) for(int i=a;i>=b;i--)
# define mct(a,b) memset(a,b,sizeof(a))
# define gc getchar()
typedef long long ll;
const int N=1e5+5;
const int inf=0x7fffffff;
const double eps=1e-7;
template <typename T> void read(T &x){
	x=0;int f=1;
	char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+c-'0';
	x*=f;
}
int n;
int main()
{
	read(n);
	printf("%lf\n",1.*((n+1)/2)/n);
	return 0;
}

T2

有n个人，每个人有一个身高
问有几个人身高大于等于k
感觉这道题比上一题简单

# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
using namespace std;

# define Rep(i,a,b) for(int i=a;i<=b;i++)
# define _Rep(i,a,b) for(int i=a;i>=b;i--)
# define mct(a,b) memset(a,b,sizeof(a))
# define gc getchar()
typedef long long ll;
const int N=2e5+5;
const int inf=0x7fffffff;
const double eps=1e-7;
template <typename T> void read(T &x){
	x=0;int f=1;
	char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+c-'0';
	x*=f;
}
int n,k,ans;
int a[N];

int main()
{
	read(n),read(k);
	Rep(i,1,n)read(a[i]);
	Rep(i,1,n)if(a[i]>=k)ans++;
	printf("%d\n",ans);
	return 0;
}

T3

给一个数组a，按$a_i$从小到大排序输出id

# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
using namespace std;

# define Rep(i,a,b) for(int i=a;i<=b;i++)
# define _Rep(i,a,b) for(int i=a;i>=b;i--)
# define mct(a,b) memset(a,b,sizeof(a))
# define gc getchar()
typedef long long ll;
const int N=2e5+5;
const int inf=0x7fffffff;
const double eps=1e-7;
template <typename T> void read(T &x){
	x=0;int f=1;
	char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+c-'0';
	x*=f;
}

int n;
struct node{
	int t,id;
	bool operator < (const node &cmp)const{
		return t<cmp.t;
	}
}a[N];

int main()
{
	read(n);
	Rep(i,1,n)read(a[i].t),a[i].id=i;
	sort(a+1,a+n+1);
	Rep(i,1,n)printf("%d ",a[i].id);
	puts("");
	return 0;
}

T4

这道题…WA了4次
细节处理很多
根据面向数据编程原则
易证：此题答案为输出n的质因数个数
证明：略

# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
using namespace std;

# define Rep(i,a,b) for(int i=a;i<=b;i++)
# define _Rep(i,a,b) for(int i=a;i>=b;i--)
# define mct(a,b) memset(a,b,sizeof(a))
# define gc getchar()
typedef long long ll;
const int N=1e5+5;
const int inf=0x7fffffff;
const double eps=1e-7;
template <typename T> void read(T &x){
	x=0;int f=1;
	char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+c-'0';
	x*=f;
}

ll a,b;
ll x;
ll ans=1;

int main()
{
	read(a),read(b);
	x=__gcd(a,b);
	for(ll i=2;i*i<=x;i++){
		if(x%i==0)ans++;
		while(x%i==0){
			x/=i;
			if(x==1)break;
		}
	}
	if(x>1)ans++;
	printf("%lld\n",ans);
	return 0;
}

T5

最为第一次写出来的T5
还是很值得纪念的
事后发现 这道题其实代码非常短
但是对于状压DP一窍不通的我还是硬着头皮写出来了
其实就是个状压的板子题
一看数据$N\le 12$,$M\le 3000$就知道了
$f_{i,j}$表示前i把钥匙开成j状态的最小花费
其实开一维就可以额，但是由于本人实在是太菜了
所以写出了一个70多行的程序…

# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
# include 
using namespace std;

# define Rep(i,a,b) for(int i=a;i<=b;i++)
# define _Rep(i,a,b) for(int i=a;i>=b;i--)
# define mct(a,b) memset(a,b,sizeof(a))
# define gc getchar()
typedef long long ll;
const int N=1005;
const int inf=0x7fffffff;
const double eps=1e-7;
template <typename T> void read(T &x){
	x=0;int f=1;
	char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f=-1;
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+c-'0';
	x*=f;
}

int n,m,high;
int ans=INT_MAX;
int a[N],b[N],c[N][20];
int f[N][5005];

inline int Qpow(int base,int ind){
	int res=1;
	while(ind){
		if(ind&1)res*=base;
		base*=base;
		ind>>=1;
	}
	return res;
}

inline int change(int x,int i){
	int num[15],cnt=0;
	memset(num,0,sizeof(num));
	while(x){
		num[++cnt]=x%2;
		x/=2;
	}
	Rep(j,1,b[i])num[c[i][j]]=1;
	int res=0;
	Rep(j,1,n)res+=num[j]*Qpow(2,j-1);
	return res;
}

int main()
{
	memset(f,0x3f,sizeof(f));
	read(n),read(m);
	high=Qpow(2,n)-1;
	f[0][0]=0;
	Rep(i,1,m){
		read(a[i]);
		read(b[i]);
		Rep(j,1,b[i])read(c[i][j]);
		Rep(j,0,high){
			f[i][j]=min(f[i][j],f[i-1][j]);
			int nxt=change(j,i);
			f[i][nxt]=min(f[i][nxt],f[i-1][j]+a[i]);
		}
	}
	Rep(i,1,m)ans=min(ans,f[i][high]);
	if(ans<1e9)printf("%d\n",ans);
	else puts("-1");
	return 0;
}