AtCoder Beginner Contest 363

A.Piling Up (\(\operatorname{Difficulty} 11\))

让你求某个数距离最近的一个 \(k\times 100\) 的距离是多少.

水.

#include<bits/stdc++.h>
using namespace std;
namespace hdk{
	namespace fastio{
		void rule(bool setting=false){std::ios::sync_with_stdio(setting);}
		inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-'){f=-1;}ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
	    inline int read(int &A){A=read();return A;}
        inline char read(char &A){A=getchar();return A;}
		inline void write(int A){if(A<0){putchar('-');A=-A;}if(A>9){write(A/10);}putchar(A%10+'0');}
        inline void write(long long A){if(A<0){putchar('-');A=-A;}if(A>9){write(A/10);}putchar(A%10+'0');}
		inline void write(char A){putchar(A);}
		inline void space(){putchar(' ');}
		inline void endl(){putchar('\n');}
		#define w(a) write(a)
		#define we(a) write(a);endl()
		#define ws(a) write(a);space()
	}
}
int power(int n,int k,int p){
	int ans,base;
	ans=1,base=n;
	while(k){
		if(k&1){
			ans*=base;
		}
		base*=base;
		k>>=1;
	}
	return ans;
}
using namespace hdk::fastio;
#define speed ios::sync_with_stdio(false);
#define tests int cases;cin>>cases;while(cases--)
#define ftests int cases=read();while(cases--)
int main(){
	int n;
	cin>>n;
	if(n<=99) cout<<99-n+1;
	else if(n<=199) cout<<199-n+1;
	else if(n<=299) cout<<299-n+1;
//	ftests{
//		
//	}
}

B.Japanese Cursed Doll (\(\operatorname{Difficulty} 32\))

显然这个题排个序直接取数组第 \(p\) 位就行了，在这里吃一发罚时太糖了，排序忘记写 less<>.

感觉这场 ABC 前面都太水了

#include<bits/stdc++.h>
using namespace std;
namespace hdk{
	namespace fastio{
		void rule(bool setting=false){std::ios::sync_with_stdio(setting);}
		inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-'){f=-1;}ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
	    inline int read(int &A){A=read();return A;}
        inline char read(char &A){A=getchar();return A;}
		inline void write(int A){if(A<0){putchar('-');A=-A;}if(A>9){write(A/10);}putchar(A%10+'0');}
        inline void write(long long A){if(A<0){putchar('-');A=-A;}if(A>9){write(A/10);}putchar(A%10+'0');}
		inline void write(char A){putchar(A);}
		inline void space(){putchar(' ');}
		inline void endl(){putchar('\n');}
		#define w(a) write(a)
		#define we(a) write(a);endl()
		#define ws(a) write(a);space()
	}
}
int power(int n,int k,int p){
	int ans,base;
	ans=1,base=n;
	while(k){
		if(k&1){
			ans*=base;
		}
		base*=base;
		k>>=1;
	}
	return ans;
}
using namespace hdk::fastio;
#define speed ios::sync_with_stdio(false);
#define tests int cases;cin>>cases;while(cases--)
#define ftests int cases=read();while(cases--)
int a[101];
bool cmp(int a,int b){
	return a>b;
}
int main(){
	int n,t,p;
	cin>>n>>t>>p;
	for(int i=1;i<=n;++i){
		cin>>a[i];
	}
	sort(a+1,a+n+1,cmp);
	int ans=max(0,t-a[p]);
	cout<<ans;
}

C.Avoid K Palindrome 2 (\(\operatorname{Difficulty} 602\))

给一个 \(S\)，求 \(S\) 的全部排列中，不包括长度为 \(k\) 的字串的有多少个.

主要是没看到数据范围，这题 \(n\le10\)，感觉再高点还有别的办法，但是既然都给这么小了那就直接爆搜吧.

第一次知道 string 能用 sort 和 next_permutation

#include<bits/stdc++.h>
using namespace std;
namespace hdk{
	namespace fastio{
		void rule(bool setting=false){std::ios::sync_with_stdio(setting);}
		inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-'){f=-1;}ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
	    inline int read(int &A){A=read();return A;}
        inline char read(char &A){A=getchar();return A;}
		inline void write(int A){if(A<0){putchar('-');A=-A;}if(A>9){write(A/10);}putchar(A%10+'0');}
        inline void write(long long A){if(A<0){putchar('-');A=-A;}if(A>9){write(A/10);}putchar(A%10+'0');}
		inline void write(char A){putchar(A);}
		inline void space(){putchar(' ');}
		inline void endl(){putchar('\n');}
		#define w(a) write(a)
		#define we(a) write(a);endl()
		#define ws(a) write(a);space()
	}
}
int power(int n,int k,int p){
	int ans,base;
	ans=1,base=n;
	while(k){
		if(k&1){
			ans*=base;
		}
		base*=base;
		k>>=1;
	}
	return ans;
}
using namespace hdk::fastio;
#define speed ios::sync_with_stdio(false);
#define tests int cases;cin>>cases;while(cases--)
#define ftests int cases=read();while(cases--)
int a[101];
bool cmp(int a,int b){
	return a>b;
}
bool check(string x,int len){
//	cout<<"check "<<x<<endl;
	for(int i=0;i+len-1<=(int)x.length()-1;++i){
		int j=i,k=i+len-1;
		bool ishw=true;
		while(j<k){
			if(x[j]!=x[k]){
//				cout<<x[j]<<" != "<<x[k]<<endl;
				ishw=false;
				break;
			}
			j++;k--;
		}
		if(ishw) return false;
	}
	return true;
}
int main(){
	int n,m;
	string x;
	cin>>n>>m>>x;
	sort(x.begin(),x.end());
	int ans=0;
	do{
		ans+=check(x,m);
	}while(next_permutation(x.begin(),x.end()));
	cout<<ans<<endl;
}

D.Palindromic Number (\(\operatorname{Difficulty} 975\))

可能是前几道里最有含金量的一个了，可惜还是道原题.

求第 \(k\) 大的回文数. 可以考虑直接预处理出一些数用来逼近

显然，对于 \(n (n \operatorname{mod} 2=0)\) 来说，\(n\) 位的回文数的个数实际上就是 \(\frac{n}{2}\) 位的自然数的个数，对于 \(n (n \operatorname{mod} 2=1)\) 的数也同理，因此可以考虑直接通过这样预处理算出第 \(k\) 大的回文数一共有几位，再然后直接按照左半部分递增逼近就行了.

#include <bits/stdc++.h>
using namespace std;
#define int long long
int n;
int a[]={0,9,9,90,90,900,900,9000,9000,90000,90000,900000,900000,9000000,9000000,90000000,90000000,900000000,900000000,9000000000,9000000000,90000000000,90000000000,900000000000,900000000000,9000000000000,9000000000000,90000000000000,90000000000000,900000000000000,900000000000000,9000000000000000,9000000000000000,90000000000000000,90000000000000000,900000000000000000,900000000000000000,9000000000000000000,9000000000000000000};
int b[100],len;
signed main(){
    int k,t,i,bk;
    cin>>n;
    if(n==1){
    	cout<<0;
    	return 0;
	}
	n--;
    len=0;
    k=1;
    while(n-a[k]>0){
		n-=a[k++];
	}
    bk=(k+1)>>1;
    t=1;
    for(i=2;i<=bk;i++){
		t*=10;
	}
    t+=n-1;
    while(t){
        b[++len]=t%10;
       	t/=10;
    }
    for(i=len;i>=1;i--){
		cout<<b[i];
	}
    for(i=k&1?2:1;i<=len;i++){
		cout<<b[i];
	}
    cout<<endl;
}

E.Sinking Land (\(\operatorname{Difficulty} 1307\))

其实就是个二维弱化版的 WaterTank，所以一开始还想着用 WaterTank 的那种单调队列去实现，后来发现数据范围能过 \(n^{2}\log n\)，所以就改了一个优先队列广搜.

优先队列广搜？这不是 DIJ?

好像是吧... 还挺一样的，但是我有一半用的深搜，其实是一样的.

#include<bits/stdc++.h>
using namespace std;
const int N=1001;
struct node{
	int x,y,w;
	bool operator <(const node &A)const{
		return w>A.w;
	}
};
int n,m,t;
priority_queue<node> q;
int a[N][N];
bool vis[N][N];
int tot;
void search(int x,int y,int t){
	if(x>1 and !vis[x-1][y]){
		if(a[x-1][y]<=t){
			vis[x-1][y]=true;
//			cout<<"Yes"<<x-1<<" "<<y<<endl;
			tot++;
			search(x-1,y,t);
		}
		else{
			q.push({x-1,y,a[x-1][y]});
		}
	}
	if(x<n and !vis[x+1][y]){
		if(a[x+1][y]<=t){
			vis[x+1][y]=true;
//			cout<<"Yes"<<x+1<<" "<<y<<endl;
			tot++;
			search(x+1,y,t);
		}
		else{
			q.push({x+1,y,a[x+1][y]});
		}
	}
	if(y>1 and !vis[x][y-1]){
		if(a[x][y-1]<=t){
			vis[x][y-1]=true;
//			cout<<"Yes"<<x<<" "<<y-1<<endl;
			tot++;
			search(x,y-1,t);
		}
		else{
			q.push({x,y-1,a[x][y-1]});
		}
	}
	if(y<m and !vis[x][y+1]){
		if(a[x][y+1]<=t){
			vis[x][y+1]=true;
//			cout<<"Yes"<<x<<" "<<y+1<<endl;
			tot++;
			search(x,y+1,t);
		}
		else{
			q.push({x,y+1,a[x][y+1]});
		}
	}
}
int main(){
	cin>>n>>m>>t;
//	cout<<"fin"<<endl;
	for(int i=1;i<=n;++i){
//		cout<<i<<endl;
		for(int j=1;j<=m;++j){
//			cout<<"input";
			cin>>a[i][j];
		} 
	}
	for(int i=1;i<=m;++i){
		q.push({1,i,a[1][i]});
		q.push({n,i,a[n][i]});
	}
	for(int i=2;i<n;++i){
		q.push({i,1,a[i][1]});
		q.push({i,m,a[i][m]});
	}
	for(int k=1;k<=t;++k){
		while(!q.empty() and q.top().w<=k){
			node u=q.top();
			q.pop();
			if(vis[u.x][u.y]) continue;
			vis[u.x][u.y]=true;
//			cout<<"Yes"<<u.x<<" "<<u.y<<endl;
			tot++;
			search(u.x,u.y,k);
		}
		cout<<n*m-tot<<endl;
	}
}

后记

纪念传奇 ABC 出 \(\operatorname{Difficulty} 3358\) 的题干掉 Tourist

Tag: ABC363