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 Everything 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\n",sum); 49 } 50 return 0; 51 }
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\n",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 }
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\n",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 }
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\n",f[n]); 21 } 22 return 0; 23 }
1011 Keen 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\n",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 }
1012 Longest 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\n",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 }