2018 Multi-University Training Contest

Solved	A	HDU 6298	Maximum Multiple
Solved	B	HDU 6299	Balanced Sequence
Solved	C	HDU 6300	Triangle Partition
Solved	D	HDU 6301	Distinct Values
	E	HDU 6302	Maximum Weighted Matching
	F	HDU 6303	Period Sequence
Solved	G	HDU 6304	Chiaki Sequence Revisited
Solved	H	HDU 6305	RMQ Similar Sequence
	I	HDU 6306	Lyndon Substring
	J	HDU 6307	Turn Off The Light
Solved	K	HDU 6308	Time Zone

A  找规律

发现只有3和4的倍数有存在答案

#include <bits/stdc++.h>
#define ll long long
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;ii++)
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")

using namespace std;
const int maxn=1e5+10;
const int maxm=1e6+10;
const int INF=0x3f3f3f3f;
ll casn,n,m,k;
#define tpyeinput ll
inline char nc() {static char buf[1000000],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;}
inline void read(tpyeinput &sum) {char ch=nc();sum=0;while(!(ch>='0'&&ch<='9')) ch=nc();while(ch>='0'&&ch<='9') sum=(sum<<3)+(sum<<1)+(ch-48),ch=nc();}
inline void read(tpyeinput &num1,tpyeinput &num2){read(num1);read(num2);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3){read(num1);read(num2);read(num3);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3,tpyeinput &num4){read(num1);read(num2);read(num3);read(num4);}
int main(){
//#define test
#ifdef test
  freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
	read(casn);
    while(casn--){
        read(n);
        if(n>=3&&n%3==0) printf("%lld
",((n/3ll)*(n/3ll)*(n/3ll)));
        else if(n>=4&&n%4==0) printf("%lld
",n*n*n/32ll);
        else puts("-1");
    }

#ifdef test
  fclose(stdin);fclose(stdout);system("out.txt");
#endif
  return 0;
}

B 贪心

去掉已经配对的,对于剩下 的,左边括号>右括号的放在左边,如果同样是左括号>右括号,不需要管哪个左括号多,只需要管哪个右括号少,右括号少的放在左边

#include <bits/stdc++.h>
#define ll long long
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;ii++)
using namespace std;
const int maxn=1e5+10;
const int maxm=1e6+10;
const int INF=0x3f3f3f3f;
int casn,n,m,k;
#define tpyeinput int
inline char nc() {static char buf[1000000],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;}
inline void read(tpyeinput &sum) {char ch=nc();sum=0;while(!(ch>='0'&&ch<='9')) ch=nc();while(ch>='0'&&ch<='9') sum=(sum<<3)+(sum<<1)+(ch-48),ch=nc();}
inline void read(tpyeinput &num1,tpyeinput &num2){read(num1);read(num2);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3){read(num1);read(num2);read(num3);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3,tpyeinput &num4){read(num1);read(num2);read(num3);read(num4);}
string str[maxn];
struct node {
	int a,b;
}s[maxn];
int cmp(node &a,node &b){
	int x=min(a.a,b.b),y=min(a.b,b.a);
	return x>y||x==y&&a.a>b.a;
}
int main(){
//#define test
#ifdef test
  freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
	IO;
	cin>>casn;
	while(casn--){
		cin>>n;
		rep(i,1,n){
			cin>>str[i];
		}
		int ans=0;
		rep(i,1,n){
			int a=0,b=0;
			rep(j,0,str[i].size()-1){
				if(str[i][j]=='(') a++;
				else if(str[i][j]==')'){
					if(a>0) a--,ans+=2;
					else b++;
				}
				s[i]=(node){a,b};
			}
		}
		sort(s+1,s+1+n,cmp);
		int a=0,b=0;
		rep(i,1,n){
			if(a>=s[i].b){
				ans+=s[i].b*2;
				a-=s[i].b;
			}else {
				ans+=a*2;
				a=0;
			}
			a+=s[i].a;
		}
		cout<<ans<<endl;
	}
#ifdef test
  fclose(stdin);fclose(stdout);system("out.txt");
#endif
  return 0;
}

C 画图,找规律

保证三点不共线,画图尝试之后发现规律,把点排序即可(这道题用输入挂反而会WA?)

#include <bits/stdc++.h>
#define ll long long
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;ii++)
using namespace std;
const int maxn=1e5+10;
const int maxm=1e6+10;
const int INF=0x3f3f3f3f;
int casn,n,m,k;
//#define tpyeinput int
//inline char nc() {static char buf[1000000],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;}
//inline void read(tpyeinput &sum) {char ch=nc();sum=0;while(!(ch>='0'&&ch<='9')) ch=nc();while(ch>='0'&&ch<='9') sum=(sum<<3)+(sum<<1)+(ch-48),ch=nc();}
//inline void read(tpyeinput &num1,tpyeinput &num2){read(num1);read(num2);}
//inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3){read(num1);read(num2);read(num3);}
//inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3,tpyeinput &num4){read(num1);read(num2);read(num3);read(num4);}
struct node {
	int x,y,id;
}p[maxn];
bool cmp(node &a,node& b){
	if(a.x==b.x) return a.y<b.y;
	return a.x<b.x;
}
int main(){
//#define test
#ifdef test
  freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
	IO;
	cin>>casn;
	while(casn--){
		cin>>n;
		rep(i,1,3*n){
			int a,b;
			cin>>a>>b;
			p[i]=(node){a,b,i};
		}
		sort(p+1,p+1+3*n,cmp);
		k=1;
		rep(i,1,n){
			rep(j,1,3){
				cout<<p[k++].id<<' ';
			}
			cout<<endl;
		}
	}
#ifdef test
  fclose(stdin);fclose(stdout);system("out.txt");
#endif
  return 0;
}

D 双指针/贪心+小根堆维护mex

排序要求的区间,然后双指针遍历所有即可,用priority_queue维护当前双指针内的mex

卡时间,建议上输入挂

#include <bits/stdc++.h>
#define ll long long
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;ii++)
using namespace std;
const int maxn=1e5+10;
const int maxm=1e6+10;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
int casn,n,m,k;
#define tpyeinput int
inline char nc() {static char buf[1000000],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;}
inline void read(tpyeinput &sum) {char ch=nc();sum=0;while(!(ch>='0'&&ch<='9')) ch=nc();while(ch>='0'&&ch<='9') sum=(sum<<3)+(sum<<1)+(ch-48),ch=nc();}
inline void read(tpyeinput &num1,tpyeinput &num2){read(num1);read(num2);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3){read(num1);read(num2);read(num3);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3,tpyeinput &num4){read(num1);read(num2);read(num3);read(num4);}
int ans[maxn];
int pos[maxn];
set<int>num;
int main(){
//#define test
#ifdef test
  freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
	read(casn);
	while(casn--){
		read(n,m);
		rep(i,1,n) {
			ans[i]=1;
			pos[i]=i;
			num.insert(i);
		}
		rep(i,1,m){
			int a,b;
			read(a,b);
			pos[b]=min(pos[b],a);
		}
		for(int i=n-1;i;i--){
			pos[i]=min(pos[i],pos[i+1]);
		}
		int last=1;
		rep(i,1,n){
			while(last<pos[i]){
				num.insert(ans[last]);
				last++;
			}
			ans[i]=*num.begin();
			num.erase(ans[i]);
		}
		rep(i,1,n){
			printf("%d%c",ans[i]," 
"[i==n]);
		}
	}

#ifdef test
  fclose(stdin);fclose(stdout);system("out.txt");
#endif
  return 0;
}

G 找规律+打表倍增

找规律发现,数列中$X$的出现次数是$lowbit(X)$,于是对于出现次数相同的值(也就是$lowbit(x)$相同)的数,构成了等差数列..

原数列就是很多个等差数列构成的了,然后先预处理一下值,每次有多少个不同的等差数列,求和即可...

具体看代码

#include <bits/stdc++.h>
#define ll long long
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<" "<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(a,b) cout<<#a<<'['<<b<<"]="<<a[b]<<endl
#define print(x) cout<<#x<<"="<<x<<' '
#define ptline() cout<<endl
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;ii++)
using namespace std;
const int maxn=1e5+10;
const int maxm=1e6+10;
const ll mod=1e9+7;
const int INF=0x3f3f3f3f;
ll casn,n,m,k;
#define tpyeinput ll
inline char nc() {static char buf[1000000],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;}
inline void read(tpyeinput &sum) {char ch=nc();sum=0;while(!(ch>='0'&&ch<='9')) ch=nc();while(ch>='0'&&ch<='9') sum=(sum<<3)+(sum<<1)+(ch-48),ch=nc();}
inline void read(tpyeinput &num1,tpyeinput &num2){read(num1);read(num2);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3){read(num1);read(num2);read(num3);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3,tpyeinput &num4){read(num1);read(num2);read(num3);read(num4);}
ll num[maxn],p[maxn];
ll inv=(mod+1)/2;
void init(){
	num[0]=p[0]=1;
	rep(i,1,63){
		num[i]=num[i-1]*2ll+1ll;
		p[i]=p[i-1]*2ll;
	}
}
ll cal(ll pos){
	ll sum=0;
	for(ll i=1;i<=pos;i*=2ll){
		ll num=(pos-i)/(2*i);
		ll m=i+num*(2ll*i);;
		num=(num+1ll)%mod;
		m=(m+i)%mod;
		m=m*num%mod;
		m=m*inv%mod;
		m=m*(__builtin_ctz(i)+1ll)%mod;
		sum=(sum+m)%mod;
	}
	return (sum+1)%mod;
}
int main(){
//#define test
#ifdef test
  freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
	read(casn);
	init();
	while(casn--){
		read(n);
		n--;
		if(!n){
			puts("1");
			continue;
		}
		ll tmp=n;
		ll pos=0;
		for(int i=62;i>=0;i--){
			if(tmp>=num[i]){
				tmp-=num[i];
				pos+=p[i];
			}
		}
		ll sum=cal(pos);
		if(tmp){
			sum=(sum+tmp%mod*(pos+1ll)%mod)%mod;
		}
		printf("%lld
",sum);
	}
#ifdef test
  fclose(stdin);fclose(stdout);system("out.txt");
#endif
  return 0;
}

H 组合数学+笛卡尔树性质

笛卡尔树的2个重要性质,

1.中序遍历等于原序列,换句话说,就是把树拍扁等价于原数列

2.区间的最大元素,就是表示了这个区间的子树的根节点

放到这个题里,就是笛卡尔树的形状和题目给的一样,问你多少种...

组合数学,dfs笛卡尔树记算总种类数,过程中需要求逆元...

代码里的这个笛卡尔树和线性推逆元是学的dls..

推导参考出题大爷:

#include <bits/stdc++.h>
#define ll long long
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;ii++)
using namespace std;
const int maxn=1e6+10;
const int maxm=1e6+10;
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
#define tpyeinput int
inline char nc() {static char buf[1000000],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;}
inline void read(tpyeinput &sum) {char ch=nc();sum=0;while(!(ch>='0'&&ch<='9')) ch=nc();while(ch>='0'&&ch<='9') sum=(sum<<3)+(sum<<1)+(ch-48),ch=nc();}
inline void read(tpyeinput &num1,tpyeinput &num2){read(num1);read(num2);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3){read(num1);read(num2);read(num3);}
inline void read(tpyeinput &num1,tpyeinput &num2,tpyeinput &num3,tpyeinput &num4){read(num1);read(num2);read(num3);read(num4);}
int casn,n,m,k;
struct node{int l,r;}ctree[maxn];
ll inv[maxn],ans;
int stk[maxn],vis[maxn];
pair<int,int> num[maxn];
#define nd ctree[now]
int dfs(int now){
	if(now==0)return 0;
	int sum=dfs(nd.l)+dfs(nd.r)+1;
	ans=ans*inv[sum]%mod;
//	cout<<ans<<endl;
	return sum;
}
void maketree(){
	int top=0;
	rep(i,1,n){
		int now=top;
		while(now&&num[stk[now-1]]>num[i]) now--;
		if(now) ctree[stk[now-1]].r=i;
		if(now<top) ctree[i].l=stk[now];
		stk[now++]=i;
		top=now;
	}
}
int main(){
//#define test
#ifdef test
  freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
	inv[1]=1;
	rep(i,2,maxn-5)inv[i]=inv[mod%i]*(mod-mod/i)%mod;
	read(casn);
	while(casn--){
		read(n);
		rep(i,1,n){
			int x;
			read(x);
			num[i]=make_pair(-x,i);
		}
		ans=inv[2]*(ll)n%mod;
//		cout<<ans<<endl;
		rep(i,1,n) ctree[i]=(node){0,0},vis[i]=0;
		maketree();
		rep(i,1,n) vis[ctree[i].l]=vis[ctree[i].r]=1;
		int rt=0;
		rep(i,1,n) if(vis[i]==0) {
			rt=i;break;
		}
		dfs(rt);
		printf("%lld
",ans);
	}
#ifdef test
  fclose(stdin);fclose(stdout);system("out.txt");
#endif
  return 0;
}

K 模拟题

难点有2个

输入,误差..

输入建议用"scanf("%d %d UTC%c%lf",&h,&m,&flag,&utc);";

　　　　也可以"cin>>h>>m>>ch>>ch>>ch>>ch>>utc";

误差就是$1.4$会被读入为$1.39999$取整变成$1.3$之类的

utc+=0.001就行了,消除误差,而且不影响答案

#include <bits/stdc++.h>
#define ll long long
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(int ii=a;ii<=b;ii++)
using namespace std;
const int maxn=1e5+10;
const int maxm=1e6+10;
const int INF=0x3f3f3f3f;
int casn,n,m,k;
int main(){
//#define test
#ifdef test
  freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);
#endif
    scanf("%d",&casn);
    int h,m,flag,ch,x;
    double y;
    while(casn--){
        y=0;
        double tt;
        scanf("%d %d UTC%c%lf",&h,&m,&flag,&tt);
        tt+=0.0001;
        x=(int)tt;
        y=tt-x;
        y*=10;
        if(flag=='-') {
        x=-(x+8);
        y=-y;}
        else x-=8;
        int t1=x;
        int t2=y*6;
        h+=t1;
        m+=t2;
        if(m>=60){
            m-=60;
            h++;
        }
        if(h>=24) h-=24;
        if(m<0){
            m+=60;
            h--;
        }if(h<0){
            h+=24;
        }
        printf("%02d:%02d
",h,m);
    }

#ifdef test
  fclose(stdin);fclose(stdout);system("out.txt");
#endif
  return 0;
}