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",((n/3ll)*(n/3ll)*(n/3ll)));

        else if(n>=4&&n%4==0) printf("%lld\n",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]," \n"[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\n",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\n",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\n",h,m);

    }

#ifdef test

  fclose(stdin);fclose(stdout);system("out.txt");

#endif

  return 0;

}

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