3.14省选模拟测试

3.14省选模拟测试

T1

自我理解

1，40分暴力$O(n^2)$
考虑状态转移方程
2，100分正解
咕~

代码

$O(n^2)$

#include<bits/stdc++.h>
#define int long long 
#define M 100099
using namespace std;
int read(){
	int f=1,re=0;
	char ch;
	for(ch=getchar();!isdigit(ch)&&ch!='-';ch=getchar());
	if(ch=='-'){f=-1,ch=getchar();}
	for(;isdigit(ch);ch=getchar()) re=(re<<3)+(re<<1)+ch-'0';
	return re*f;
}
int n,m,f[M],mul[M],d[M];
const int mod=998244353; 
int ksm(int a,int b){
	if(a==0||a==1) return 1;
    int ans=1;
    while(b){
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }return ans%mod;
}
int C(int x,int y){
	return mul[x]*d[x-y]%mod*d[y]%mod;
}
signed main(){
	//freopen("forest.in","r",stdin);
	//freopen("forest.out","w",stdout);
	n=read(),m=read();
	mul[0]=f[0]=d[0]=d[1]=1;
	for(int i=1;i<=n;i++) mul[i]=mul[i-1]*i%mod;
	for(int i=2;i<=n;i++) d[i]=d[mod%i]*(mod-mod/i)%mod;
	for(int i=2;i<=n;i++) d[i]=d[i-1]*d[i]%mod;
	for(int i=1;i<=n;i++)
		for(int j=i-1;j>=0;j--){
			if(i-j>m) break;
			else f[i]=(f[i]+(f[j]*C(i-1,j)%mod*ksm(i-j,i-j-2)%mod))%mod;
		}printf("%lld\n",f[n]);
	return 0; 
} 

$O(nlogn)$

T2

自我理解

1，30分 暴力
枚举左端点，再枚举长度，然后判断循环节，复杂度$O(\frac{n^2}{k})$
2，100分正解
咕~

代码

$O(\frac{n^2}{k})$

#include<bits/stdc++.h>
#define LL unsigned long long
#define M 300009
using namespace std;
LL mi[M],has[M],n,k;
const LL h=37;
int ans;
char s[M];
int read(){
	int f=1,re=0;
	char ch;
	for(ch=getchar();!isdigit(ch)&&ch!='-';ch=getchar());
	if(ch=='-'){f=-1,ch=getchar();}
	for(;isdigit(ch);ch=getchar()) re=(re<<3)+(re<<1)+ch-'0';
	return re*f;
}
LL gethas(int l,int r){return has[r]-has[l-1]*mi[r-l+1];}
void work1(){
	mi[0]=1;
	for(int i=1;i<=n;i++) mi[i]=mi[i-1]*h;
	for(int i=1;i<=n;i++) has[i]=has[i-1]*h+(s[i]-'a');
	for(int i=1;i<=n;i++)
		for(int j=k;j<=n;j+=k){
			int l=i,r=l+j-1,len=j/k;
			if(gethas(l,r-len)==gethas(l+len,r)) ans++;
		}printf("%d\n",ans);
}
int main(){
	freopen("sutoringu.in","r",stdin);
	freopen("sutoringu.out","w",stdout);
	n=read(),k=read();
	scanf("%s",s+1);
	if((n*n/k)<1e8+9) work1();
	return 0; 
} 

$O(nlo{g}^{2}n)$

T3

自我理解

正解：枚举直角边，然后计算出第三个点，用哈希表判断存在性，细节较多需要注意。
（本题数据没有重点，但该代码考虑了重点）

代码

#include<bits/stdc++.h>
#define int long long 
#define M 10009
using namespace std;
int read(){
	int f=1,re=0;
	char ch;
	for(ch=getchar();!isdigit(ch)&&ch!='-';ch=getchar());
	if(ch=='-'){f=-1,ch=getchar();}
	for(;isdigit(ch);ch=getchar()) re=(re<<3)+(re<<1)+ch-'0';
	return re*f;
}
struct data{
	int x,y,nxt,cnt;
}node[M];
const int mod=1e4+7;
int a[M],b[M],n,ans,has[M*100],cur=1,num;
int getdis(int x,int y){return (a[x]-a[y])*(a[x]-a[y])+(b[x]-b[y])*(b[x]-b[y]);} 
bool check(int x,int y,int z){
	int lx=getdis(x,y),ly=getdis(x,z),lz=getdis(y,z);
	if(lx==ly&&lx+ly==lz) return 1;
	if(lx==lz&&lx+lz==ly) return 1;
	if(lz==ly&&lz+ly==lx) return 1;
	return 0;
}
void work1(){
	for(int i=1;i<=n;i++)
		for(int j=i+1;j<=n;j++)
			for(int k=j+1;k<=n;k++)
				if(check(i,j,k)) ans++;
	printf("%lld\n",ans);
}
void insert(int x,int y){  
    int h=(x*x+y*y)%mod;  
    node[cur].x=x;  
    node[cur].y=y;  
    node[cur].nxt=has[h]; 
	node[cur].cnt=1; 
    has[h]=cur++;  
}  
int judge(int x,int y)  {  
    int h=(x*x+y*y)%mod;    
    int nxt=has[h];  
    while(nxt!=-1){  
        if(x==node[nxt].x&&y==node[nxt].y) return nxt;  
        nxt=node[nxt].nxt;  
    }return 0;  
}
void work2(){
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++){
			if(i==j) continue;
			if(a[i]==a[j]&&b[i]==b[j]) continue;
			int x1=a[i]+b[i]-b[j]; 
			int y1=b[i]+a[j]-a[i];
			//printf("%lld %lld %lld %lld %lld %lld\n",a[i],b[i],a[j],b[j],x1,y1);
			if(num=judge(x1,y1)) ans+=node[num].cnt;
			x1=a[i]-b[i]+b[j];
			y1=b[i]+a[i]-a[j];
			if(num=judge(x1,y1)) ans+=node[num].cnt;
			//printf("%lld %lld %lld %lld %lld %lld\n",a[i],b[i],a[j],b[j],x1,y1);
		}//printf("%lld\n",ans);
		printf("%lld\n",ans/2);
	return;	
}
signed main(){
	freopen("triangle.in","r",stdin);
	freopen("triangle.out","w",stdout);
	n=read();
	memset(has,-1,sizeof(has));
	for(int i=1;i<=n;i++){
		a[i]=read(),b[i]=read();
		if(num=judge(a[i],b[i])) node[num].cnt++;
		else insert(a[i],b[i]);
	}if(n<=300) work1();
	else work2();
	return 0; 
}