zoukankan html css js c++ java

2019杭电多校第二场

比赛总结

@辣鸡lfw，一开始算三角形那题没想到正解使用莫队乱搞，然后要选出50个最大的时候还用莫队。。。主席树区间第k大什么的完全忘记了

题解

1001 Another Chess Problem

unsolved

1002 Beauty Of Unimodal Sequence

unsolved

1003 Coefficient

unsolved

1004 Double Tree

unsolved

1005 Everying is Generated In Equal Probability

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define mod 998244353
 5 const int maxn=3000;
 6 const int maxm=3010;
 7 int n;
 8 ll sum,f[maxm],rav[maxm],b[maxm];
 9 ll Pow(ll x,int n)
10 {
11     ll ans=1;
12     while(n)
13     {
14         if(n&1)ans=ans*x%mod;
15         n>>=1;
16         x=x*x%mod;
17     }
18     return ans;
19 }
20 ll C(int a,int b){return (f[a]*rav[b]%mod)*rav[a-b]%mod;}
21 void Init()
22 {
23     sum=0; f[0]=rav[0]=1;
24     for(ll i=1;i<=maxn;i++)f[i]=f[i-1]*i%mod;
25     rav[maxn]=Pow(f[maxn],mod-2);
26     for(ll i=maxn-1;i;i--) rav[i]=rav[i+1]*(i+1)%mod;
27 }
28 /////////// 求组合数和逆元 
29 
30 int main()
31 { 
32     while(~scanf("%d",&n)) 
33     {
34         Init();
35         
36         ll r=Pow(4,mod-2),k,x,y;
37         for(ll i=2;i<=n;++i)
38         {
39             x=Pow(2,i);
40             y=Pow(Pow(2,i)-1,mod-2);
41             k=x*y%mod;
42             b[i]=k*((i*(i-1)%mod)*r%mod)%mod;
43             for(int j=0;j<i;++j) b[i]=(b[i]+(y*C(i,j)%mod)*b[j]%mod)%mod;
44             sum=(sum+b[i])%mod;
45         }
46         sum=sum*Pow(n,mod-2)%mod;
47         
48         printf("%lld
",sum);
49     }
50     return 0;
51 }

View Code

1006 Fantastic Magic Cube

unsolved

1007 Game

unsolved

1008 Harmonious Army

题解：https://blog.csdn.net/baiyifeifei/article/details/97231963

 1 #include<bits/stdc++.h>
 2 #define maxl 200010
 3 using namespace std;
 4  
 5 int n,q,cnt,tot;
 6 int a[maxl],num[maxl],rt[maxl];
 7 long long b[110];
 8 struct node
 9 {
10     int ls,rs,sum;
11 }tree[maxl*40];
12  
13 inline void insert(int num,int &x,int l,int r)
14 {
15     tree[++tot]=tree[x];x=tot;
16     ++tree[x].sum;
17     if(l==r) return;
18     int mid=(l+r)>>1;
19     if(num<=mid)
20         insert(num,tree[x].ls,l,mid);
21     else
22         insert(num,tree[x].rs,mid+1,r);
23 }
24  
25 inline int query(int i,int j,int k,int l,int r)
26 {
27     if(l==r)
28         return l;
29     int tp=tree[tree[j].ls].sum-tree[tree[i].ls].sum;
30     int mid=(l+r)>>1;
31     if(k<=tp) 
32         return query(tree[i].ls,tree[j].ls,k,l,mid);
33     else
34         return query(tree[i].rs,tree[j].rs,k-tp,mid+1,r);
35 }
36  
37 inline void prework()
38 {
39     for(int i=1;i<=n;i++)
40         scanf("%d",&a[i]),num[i]=a[i];
41     sort(num+1,num+1+n);
42     cnt=unique(num+1,num+1+n)-num-1;
43     for(int i=1;i<=n;i++)
44         a[i]=lower_bound(num+1,num+1+cnt,a[i])-num;
45     for(int i=0;i<=tot;i++)
46         tree[i].ls=tree[i].rs=tree[i].sum=0;
47     tot=0;
48     for(int i=1;i<=n;i++)
49         rt[i]=0;
50     for(int i=1;i<=n;i++)
51     {
52         rt[i]=rt[i-1];
53         insert(a[i],rt[i],1,cnt);
54     }
55 }
56  
57 inline void mainwork()
58 {
59     int l,r,up;
60     long long ans=0;
61     for(int i=1;i<=q;i++)
62     {
63         scanf("%d%d",&l,&r);
64         up=min(50,r-l+1);
65         for(int j=1;j<=up;j++)
66             b[j]=num[query(rt[l-1],rt[r],r-l+1-j+1,1,cnt)];
67         ans=-1;
68         for(int j=1;j<=up-2;j++)
69         if(b[j]<b[j+1]+b[j+2])
70         {
71             ans=b[j]+b[j+1]+b[j+2];
72             break;
73         }
74         printf("%lld
",ans);
75     }
76 }
77  
78 int main()
79 {
80     while(~scanf("%d%d",&n,&q))
81     {
82         prework();
83         mainwork();
84         //print();
85     }
86     return 0;
87 }

View Code

1009 I Love Palindrome String

 1 #include<bits/stdc++.h>
 2 #define maxl 200010
 3 using namespace std;
 4  
 5 int n,q,cnt,tot;
 6 int a[maxl],num[maxl],rt[maxl];
 7 long long b[110];
 8 struct node
 9 {
10     int ls,rs,sum;
11 }tree[maxl*40];
12  
13 inline void insert(int num,int &x,int l,int r)
14 {
15     tree[++tot]=tree[x];x=tot;
16     ++tree[x].sum;
17     if(l==r) return;
18     int mid=(l+r)>>1;
19     if(num<=mid)
20         insert(num,tree[x].ls,l,mid);
21     else
22         insert(num,tree[x].rs,mid+1,r);
23 }
24  
25 inline int query(int i,int j,int k,int l,int r)
26 {
27     if(l==r)
28         return l;
29     int tp=tree[tree[j].ls].sum-tree[tree[i].ls].sum;
30     int mid=(l+r)>>1;
31     if(k<=tp) 
32         return query(tree[i].ls,tree[j].ls,k,l,mid);
33     else
34         return query(tree[i].rs,tree[j].rs,k-tp,mid+1,r);
35 }
36  
37 inline void prework()
38 {
39     for(int i=1;i<=n;i++)
40         scanf("%d",&a[i]),num[i]=a[i];
41     sort(num+1,num+1+n);
42     cnt=unique(num+1,num+1+n)-num-1;
43     for(int i=1;i<=n;i++)
44         a[i]=lower_bound(num+1,num+1+cnt,a[i])-num;
45     for(int i=0;i<=tot;i++)
46         tree[i].ls=tree[i].rs=tree[i].sum=0;
47     tot=0;
48     for(int i=1;i<=n;i++)
49         rt[i]=0;
50     for(int i=1;i<=n;i++)
51     {
52         rt[i]=rt[i-1];
53         insert(a[i],rt[i],1,cnt);
54     }
55 }
56  
57 inline void mainwork()
58 {
59     int l,r,up;
60     long long ans=0;
61     for(int i=1;i<=q;i++)
62     {
63         scanf("%d%d",&l,&r);
64         up=min(50,r-l+1);
65         for(int j=1;j<=up;j++)
66             b[j]=num[query(rt[l-1],rt[r],r-l+1-j+1,1,cnt)];
67         ans=-1;
68         for(int j=1;j<=up-2;j++)
69         if(b[j]<b[j+1]+b[j+2])
70         {
71             ans=b[j]+b[j+1]+b[j+2];
72             break;
73         }
74         printf("%lld
",ans);
75     }
76 }
77  
78 int main()
79 {
80     while(~scanf("%d%d",&n,&q))
81     {
82         prework();
83         mainwork();
84         //print();
85     }
86     return 0;
87 }

View Code

1010 Just Skip The Problem

就直接问n个2进制位上的数，由于有顺序，那么就是n!，对1e6+3取模，那就在外面预处理到1e6+3，之后都是0

 1 #include<bits/stdc++.h>
 2 #define maxl 1000010
 3 using namespace std;
 4 
 5 const int mod=1e6+3;
 6 
 7 long long f[maxl]; 
 8 
 9 int main()
10 {
11     f[1]=1;
12     for(int i=2;i<mod;i++)
13         f[i]=(f[i-1]*i)%mod;
14     int n;
15     while(~scanf("%d",&n))
16     {
17         if(n>=mod)
18             puts("0");
19         else
20             printf("%lld
",f[n]);
21     }
22     return 0;
23 }

View Code

1011 Keep On Everything But TRiangle

题解：https://blog.csdn.net/liufengwei1/article/details/97324278

 1 #include<bits/stdc++.h>
 2 #define maxl 200010
 3 using namespace std;
 4  
 5 int n,q,cnt,tot;
 6 int a[maxl],num[maxl],rt[maxl];
 7 long long b[110];
 8 struct node
 9 {
10     int ls,rs,sum;
11 }tree[maxl*40];
12  
13 inline void insert(int num,int &x,int l,int r)
14 {
15     tree[++tot]=tree[x];x=tot;
16     ++tree[x].sum;
17     if(l==r) return;
18     int mid=(l+r)>>1;
19     if(num<=mid)
20         insert(num,tree[x].ls,l,mid);
21     else
22         insert(num,tree[x].rs,mid+1,r);
23 }
24  
25 inline int query(int i,int j,int k,int l,int r)
26 {
27     if(l==r)
28         return l;
29     int tp=tree[tree[j].ls].sum-tree[tree[i].ls].sum;
30     int mid=(l+r)>>1;
31     if(k<=tp) 
32         return query(tree[i].ls,tree[j].ls,k,l,mid);
33     else
34         return query(tree[i].rs,tree[j].rs,k-tp,mid+1,r);
35 }
36  
37 inline void prework()
38 {
39     for(int i=1;i<=n;i++)
40         scanf("%d",&a[i]),num[i]=a[i];
41     sort(num+1,num+1+n);
42     cnt=unique(num+1,num+1+n)-num-1;
43     for(int i=1;i<=n;i++)
44         a[i]=lower_bound(num+1,num+1+cnt,a[i])-num;
45     for(int i=0;i<=tot;i++)
46         tree[i].ls=tree[i].rs=tree[i].sum=0;
47     tot=0;
48     for(int i=1;i<=n;i++)
49         rt[i]=0;
50     for(int i=1;i<=n;i++)
51     {
52         rt[i]=rt[i-1];
53         insert(a[i],rt[i],1,cnt);
54     }
55 }
56  
57 inline void mainwork()
58 {
59     int l,r,up;
60     long long ans=0;
61     for(int i=1;i<=q;i++)
62     {
63         scanf("%d%d",&l,&r);
64         up=min(50,r-l+1);
65         for(int j=1;j<=up;j++)
66             b[j]=num[query(rt[l-1],rt[r],r-l+1-j+1,1,cnt)];
67         ans=-1;
68         for(int j=1;j<=up-2;j++)
69         if(b[j]<b[j+1]+b[j+2])
70         {
71             ans=b[j]+b[j+1]+b[j+2];
72             break;
73         }
74         printf("%lld
",ans);
75     }
76 }
77  
78 int main()
79 {
80     while(~scanf("%d%d",&n,&q))
81     {
82         prework();
83         mainwork();
84         //print();
85     }
86     return 0;
87 }

View Code

1012 Longgest Subarray

题解：https://blog.csdn.net/liufengwei1/article/details/97324937

 1 #include<bits/stdc++.h>
 2 #define maxl 200010
 3 using namespace std;
 4  
 5 int n,q,cnt,tot;
 6 int a[maxl],num[maxl],rt[maxl];
 7 long long b[110];
 8 struct node
 9 {
10     int ls,rs,sum;
11 }tree[maxl*40];
12  
13 inline void insert(int num,int &x,int l,int r)
14 {
15     tree[++tot]=tree[x];x=tot;
16     ++tree[x].sum;
17     if(l==r) return;
18     int mid=(l+r)>>1;
19     if(num<=mid)
20         insert(num,tree[x].ls,l,mid);
21     else
22         insert(num,tree[x].rs,mid+1,r);
23 }
24  
25 inline int query(int i,int j,int k,int l,int r)
26 {
27     if(l==r)
28         return l;
29     int tp=tree[tree[j].ls].sum-tree[tree[i].ls].sum;
30     int mid=(l+r)>>1;
31     if(k<=tp) 
32         return query(tree[i].ls,tree[j].ls,k,l,mid);
33     else
34         return query(tree[i].rs,tree[j].rs,k-tp,mid+1,r);
35 }
36  
37 inline void prework()
38 {
39     for(int i=1;i<=n;i++)
40         scanf("%d",&a[i]),num[i]=a[i];
41     sort(num+1,num+1+n);
42     cnt=unique(num+1,num+1+n)-num-1;
43     for(int i=1;i<=n;i++)
44         a[i]=lower_bound(num+1,num+1+cnt,a[i])-num;
45     for(int i=0;i<=tot;i++)
46         tree[i].ls=tree[i].rs=tree[i].sum=0;
47     tot=0;
48     for(int i=1;i<=n;i++)
49         rt[i]=0;
50     for(int i=1;i<=n;i++)
51     {
52         rt[i]=rt[i-1];
53         insert(a[i],rt[i],1,cnt);
54     }
55 }
56  
57 inline void mainwork()
58 {
59     int l,r,up;
60     long long ans=0;
61     for(int i=1;i<=q;i++)
62     {
63         scanf("%d%d",&l,&r);
64         up=min(50,r-l+1);
65         for(int j=1;j<=up;j++)
66             b[j]=num[query(rt[l-1],rt[r],r-l+1-j+1,1,cnt)];
67         ans=-1;
68         for(int j=1;j<=up-2;j++)
69         if(b[j]<b[j+1]+b[j+2])
70         {
71             ans=b[j]+b[j+1]+b[j+2];
72             break;
73         }
74         printf("%lld
",ans);
75     }
76 }
77  
78 int main()
79 {
80     while(~scanf("%d%d",&n,&q))
81     {
82         prework();
83         mainwork();
84         //print();
85     }
86     return 0;
87 }

View Code

查看全文

相关阅读:
oracle 10g 免安装客户端在windows下配置
 sql2005 sa密码
 使用windows live writer 有感
 windows xp SNMP安装包提取
 汉化groove2007
迁移SQL server 2005 Reporting Services到SQL server 2008 Reporting Services全程截图操作指南
 foxmail 6在使用中的问题
 AGPM客户端连接不上服务器解决一例
 SpringSource Tool Suite add CloudFoundry service
Java 之 SWing

原文地址：https://www.cnblogs.com/csushl/p/11249771.html