Too much food can only participate in the TG group
PJ next door than the TG test of how hard it is fat thing? ?
Egg explosion process
8:20 to get up
Then load side while other issues yeah breakfast,What a mess this server ccf
8:40 only to see the face of questions, first read it again all ABC
A50 minutes played first, then think about B (9:00)
B launched a push for a long time to exchange a maximum of only affect the adjacent reverse the conclusion of a number, but still thought the 40 minute practice, went back to see A
A 60 minutes and then found a parity of sentence on the line (do not know right), then one will hit finished, (9:30
And then went to push B, blind push even launch a conclusion: (9:50
Defined \ (S_I \) for the first \ (I \) before than the first number (I \) \ large number number, then each operation are all \ (S_I \) forward movement and a subtracted 1,0 without reduction
It is to query a number greater than the number and range and the number of period
I thought this is a dynamic two-dimensional problem of the number of points, then keep trying to write a two-dimensional BIT Han
At 10:50, when I suddenly thought, \ (S_1 \) must be 0, then it becomes a number greater than the entire array query within a certain number and the number and
right! value! line! segment! tree!
11:00 After a sample, 11:10 to complete the shoot
For each k then C is equivalent to title number n of the same size consisting of several rings, can enumerate factor, an analog answer violence
In 11:59:43 done to pay up
But there is a place to write the bombing, it may burst egg / kk
Desirable Score: 60 + 100 + 100 = 260
I have heard that C class in addition to the AK QAQ / kk %%%%%%%%%%%%%%%%%%%%%%%%
answer
A
1 to 5 for the direct determination \ (a_1-b_1 \) is equal to \ (a_2-b_2 \) to
For the 6 to 10, it is judged \ (a_1-b_1 = b_2- a_2 \) or \ (A_2-B_2 = B_1-A_1 \) , a two to satisfy
For 11 to 12, if the \ (t = 1 \) and \ (t = 2 \) operation wherein only a direct do so in accordance with the practice of 1 to 10
Otherwise, you can determine what parity judgment \ ((a_1-b_1) \ bmod 2 = (a_2-b_2) \ bmod 2 \) to
The rest will not be expected to score 60
// This code wrote by chtholly_micromaker(MicroMaker)
#include <bits/stdc++.h>
#define reg register
#define int long long
using namespace std;
const int MaxN=100050;
struct Quetion
{
int t,u,v;
}Q[MaxN];
template <class t> inline void read(t &s)
{
s=0;
reg int f=1;
reg char c=getchar();
while(!isdigit(c))
{
if(c=='-')
f=-1;
c=getchar();
}
while(isdigit(c))
s=(s<<3)+(s<<1)+(c^48),c=getchar();
s*=f;
return;
}
int n,m;
int a[MaxN],b[MaxN];
namespace SUBTASK1
{
inline void solve()
{
if(a[1]-b[1]==a[2]-b[2])
puts("YES");
else
puts("NO");
return;
}
}
namespace SUBTASK2
{
inline void solve()
{
if((a[1]-b[1]==b[2]-a[2])||(a[2]-b[2]==b[1]-a[1]))
puts("YES");
else
puts("NO");
return;
}
}
namespace SUBTASK3
{
inline void solve(int c1,int c2)
{
if(c1&&c2)
{
if(((a[1]-b[1])&1)==((a[2]-b[2])&1))
puts("YES");
else
puts("NO");
return;
}
else
{
if(c1)
SUBTASK1::solve();
else
SUBTASK2::solve();
}
return;
}
}
inline void work()
{
read(n);read(m);
for(int i=1;i<=n;++i)
read(a[i]);
for(int i=1;i<=n;++i)
read(b[i]);
reg int cnt1=0,cnt2=0;
for(int i=1;i<=m;++i)
{
read(Q[i].t);read(Q[i].u);read(Q[i].v);
if(Q[i].t==2)
++cnt2;
if(Q[i].t==1)
++cnt1;
}
if(n==2&&m==1&&cnt1)
{
SUBTASK1::solve();
return;
}
if(n==2&&m==1&&cnt2)
{
SUBTASK2::solve();
return;
}
if(n==2)
{
SUBTASK3::solve(cnt1,cnt2);
return;
}
return;
}
signed main(void)
{
freopen("sequence.in","r",stdin);
freopen("sequence.out","w",stdout);
int t;cin>>t;
while(t--)
work();
return 0;
}B
We prosequence set to \ (a_i \) , \ (S_I \) of \ (a_1, a_2, \ cdots , a_ {i-1} \) is greater than \ (a_i \) number of the number of
For a \ (A \) sequence requirements \ (S \) sequence may be used Fenwick tree \ (O (n \ log n ) \) implemented
Let's look at an example (assuming that \ (s_i \) has been seeking good)
a: 1 4 3 5 2
s: 0 0 1 0 3After bubbling round
a: 1 3 4 2 5
s: 0 0 0 2 0One will find that each is greater than 0 \ (s_i \) are subtracted 1 and all \ (s_i \) are to the left of a
Then you will find, \ (S_1 \) certain to 0 (by the \ (s_i \) of the definition of shows)
Then left shift operation can be ignored, then \ (t = 2 \) operation becomes
Current demand \ (A \) sequence \ (\ sum_ {i = 1 } ^ n \ max (s_i-c, 0) \)
It is asked (in this case refers to the sequence \ (S \) sequence)
\ [\ text {all particles larger than the sequence} c \ text {numbers and} - \ text {greater than the sequence} c \ text {number th number} \ times c \]
this direct weight maintenance on the line segment tree (tree or two arrays may be)
Now, if the switching operation is re-evaluated by direct violence \ (S \) sequence able to get the 60 minutes
But not far away from the 100 points
We take a look at:
Suppose \ (p <q \)
a: p q
s: x yExchange \ (P, Q \) , since \ (p <q \) so \ (P \) corresponding to \ (S \) to \ (+ 1 \) , and \ (Q \) corresponding to \ (S \ ) will not change
a: q p
s: y x+1\ (p> q \) almost
Therefore we use this exchange operation as long as the properties of each single point modify \ (S \) on the line, a single modification complexity \ (O (\ log n) \)
The total time complexity \ (O (m \ log n ) \)
// This code wrote by chtholly_micromaker(MicroMaker)
#include <bits/stdc++.h>
#define reg register
#define int long long
using namespace std;
const int MaxN=200050;
template <class t> inline void read(t &s)
{
s=0;
reg int f=1;
reg char c=getchar();
while(!isdigit(c))
{
if(c=='-')
f=-1;
c=getchar();
}
while(isdigit(c))
s=(s<<3)+(s<<1)+(c^48),c=getchar();
s*=f;
return;
}
int n,m;
int a[MaxN],b[MaxN];
int tr[MaxN];
int val[MaxN<<2];
int sum[MaxN<<2];
inline int lowbit(int x)
{
return x&(-x);
}
inline void modify(int x,int k)
{
while(x<=n)
{
tr[x]+=k;
x+=lowbit(x);
}
return;
}
inline int queryP(int x)
{
reg int res=0;
while(x>0)
{
res+=tr[x];
x-=lowbit(x);
}
return res;
}
inline int query(int l,int r)
{
return queryP(r)-queryP(l-1);
}
#define lson (u<<1)
#define rson (u<<1|1)
inline void Spushup(int u)
{
val[u]=val[lson]+val[rson];
sum[u]=sum[lson]+sum[rson];
return;
}
inline void Smodify(int u,int l,int r,int p,int k)
{
if(l==r)
{
// printf("Smodify %lld %lldx%lld\n",u,k,l);
val[u]+=k;
sum[u]+=k*l;
return;
}
reg int mid=(l+r)>>1;
if(p<=mid)
Smodify(lson,l,mid,p,k);
else
Smodify(rson,mid+1,r,p,k);
Spushup(u);
return;
}
inline pair<int,int> Squery(int u,int l,int r,int ql,int qr)
{
if(ql<=l&&r<=qr)
return make_pair(val[u],sum[u]);
reg int mid=(l+r)>>1;
reg pair<int,int> res=make_pair(0,0);
if(ql<=mid)
{
reg pair<int,int> t=Squery(lson,l,mid,ql,qr);
res.first+=t.first;
res.second+=t.second;
}
if(mid<qr)
{
reg pair<int,int> t=Squery(rson,mid+1,r,ql,qr);
res.first+=t.first;
res.second+=t.second;
}
// printf("Squery %lld %lld is %lld %lld\n",l,r,res.first,res.second);
return res;
}
inline void recalc()
{
// memset(tr,0,sizeof tr);
// memset(val,0,sizeof val);
// memset(sum,0,sizeof sum);
for(int i=1;i<=n;++i)
{
b[i]=query(a[i],n);
// printf("recalc %lld is %lld\n",i,t);
if(b[i])
Smodify(1,1,n,b[i],1);
modify(a[i],1);
}
return;
}
inline void work(int i,int j)
{
if(a[i]>a[j])
{
if(b[j]>0)
Smodify(1,1,n,b[j],-1);
--b[j];
if(b[j]>0)
Smodify(1,1,n,b[j],1);
swap(a[i],a[j]);
swap(b[i],b[j]);
}
else
{
if(b[i]>0)
Smodify(1,1,n,b[i],-1);
++b[i];
if(b[i]>0)
Smodify(1,1,n,b[i],1);
swap(a[i],a[j]);
swap(b[i],b[j]);
}
return;
}
signed main(void)
{
freopen("bubble.in","r",stdin);
freopen("bubble.out","w",stdout);
cin>>n>>m;
for(int i=1;i<=n;++i)
read(a[i]);
recalc();
reg int t,c;
for(int i=1;i<=m;++i)
{
read(t);read(c);
if(t==1)
{
// swap(a[c],a[c+1]);
work(c,c+1);
}
else
{
if(c>=n)
puts("0");
else
{
reg pair<int,int> res=Squery(1,1,n,c+1,n);
// printf("query %lld %lld is %lld %lld\n",c+1,n,res.first,res.second);
printf("%lld\n",res.second-res.first*c);
}
}
}
return 0;
}C
To play with the strong hand, length of the loop is found \ (\ frac n {\ gcd (n, k)} \)
The answer must be the same length and the same ring, the factors enumerated directly, pretreatment
\ (\ {1,2,3,4,5,6 \} \) in \ (k = 1 \) when put into the \ (\ {2,4,6,5,3,1 \} \)
Can be found, from largest to smallest is the case, first put 6, 6 put five on the right, left 6 put the right place on the right of the left 4,6 3,6 2 to put the left
So obviously, to get two hands on the line
Then \ (k> 1 \) when such a process can be performed on each ring
Remember in particular deal with what \ (k = 0 \) case
// This code wrote by chtholly_micromaker(MicroMaker)
#include <bits/stdc++.h>
#define reg register
#define int long long
using namespace std;
const int MaxN=200050;
template <class t> inline void read(t &s)
{
s=0;
reg int f=1;
reg char c=getchar();
while(!isdigit(c))
{
if(c=='-')
f=-1;
c=getchar();
}
while(isdigit(c))
s=(s<<3)+(s<<1)+(c^48),c=getchar();
s*=f;
return;
}
int n;
int a[MaxN],b[MaxN],ans[MaxN];
int thezero;
inline int dfs(int l,int r)
{
// printf("dfs %lld %lld\n",l,r);
reg int L=1e5;
reg int R=1e5-1;
for(int i=r;i>=l;--i)
if(i&1)
b[++R]=a[i];
else
b[--L]=a[i];
reg int res=0;
for(int i=L;i<=R;++i)
if(i==R)
res+=b[R]*b[L];
else
res+=b[i]*b[i+1];
// puts("ok");
return res;
}
inline int gcd(int a,int b)
{
return !b?a:gcd(b,a%b);
}
signed main(void)
{
// freopen("ring.in","r",stdin);
// freopen("ring.out","w",stdout);
int m;
cin>>n>>m;
for(int i=1;i<=n;++i)
read(a[i]),thezero+=a[i]*a[i];
sort(a+1,a+n+1);
for(int i=1;i<n;++i)
if(!(n%i))
{
for(int j=1;j<=n;j+=n/i)
ans[i]+=dfs(j,j+n/i-1);
}
for(int i=1;i<=m;++i)
{
reg int c;
read(c);
if(!c)
printf("%lld\n",thezero);
else
printf("%lld\n",ans[gcd(c,n)]);
}
return 0;
}