Questions surface: https://www.cnblogs.com/Juve/articles/11599318.html
antipalindome:
Hit the table to find the law?
For a palindrome string, as long as we do not within three palindromes can be, that they do not appear within three to legitimate palindrome
For the first three: m * (m-1) * (m-2), the remaining n-3 positions with m-2 to fill
Therefore $ ans = m * (m-1) * (m-2) ^ (n-2) $, noted that the boundary is determined
#include<iostream>
#include<cstdio>
#include<cstring>
#define int long long
using namespace std;
const int mod=1e9+7;
const int phi=1e9+6;
int t,n,m,ans=0;
int q_pow(int a,int b,int p){
a%=p;
(b+=(p-1))%=(p-1);
int res=1;
while(b){
if(b&1) res=res*a%p;
a=a*a%p;
b>>=1;
}
return res%p;
}
signed main(){
scanf("%lld",&t);
while(t--){
ans=0;
scanf("%lld%lld",&n,&m);
if(n==1){
printf("%lld\n",m%mod);
continue;
}else if(m==1){
puts("0");
continue;
}else if(m==2&&n==2){
puts("2");
continue;
}else{
ans=q_pow(m-2,n-2,mod)%mod;
for(int i=m;i>=m-1;--i){
(ans*=(i%mod))%=mod;
}
printf("%lld\n",ans);
}
}
return 0;
}
randomwalking：
Tree dp, feel it troublesome thief, a blogger is not
#include<iostream>
#include<cstdio>
#include<cstring>
#define int long long
using namespace std;
const int MAXN=1e6+4;
const double inf=110000000000000000.0;
int n,a[MAXN],du[MAXN],ans;
int to[MAXN<<1],nxt[MAXN<<1],pre[MAXN],cnt=0;
void add(int u,int v){
++cnt,to[cnt]=v,nxt[cnt]=pre[u],pre[u]=cnt;
}
double f[MAXN],g[MAXN],minn=inf;
void dfs(int x,int fa){
f[x]=(double)a[x];
if(du[x]==1&&fa!=0){
return ;
}
for(int i=pre[x];i;i=nxt[i]){
int y=to[i];
if(y==fa) continue;
dfs(y,x);
if(fa!=0&&du[x]!=1) f[x]+=(double)f[y]/(double)((double)du[x]-1.0);
else f[x]+=(double)f[y]/du[x];
}
}
void DFS(int x,int fa){
for(int i=pre[x];i;i=nxt[i]){
int y=to[i];
if(y==fa) continue;
g[y]=(double)a[y];
double tmp1=(double)(g[x]-(double)a[x])*(double)du[x];
double tmp2=tmp1-f[y];
double tmp3=(double)(f[y]-(double)a[y])*(double)(du[y]-1.0);
if(du[x]>1) g[y]+=(double)(tmp2/(double)(du[x]-1.0)+(double)a[x]+tmp3)/(double)(du[y]);
else g[y]+=(double)((double)a[x]+tmp3)/(double)(du[y]);
DFS(y,x);
}
}
signed main(){
scanf("%lld",&n);
for(int i=1;i<=n;++i){
scanf("%lld",&a[i]);
}
for(int i=1,u,v;i<n;++i){
scanf("%lld%lld",&u,&v);
add(u,v),add(v,u);
++du[u],++du[v];
}
dfs(1,0);
g[1]=f[1];
DFS(1,0);
minn=g[1],ans=1;
for(int i=1;i<=n;++i){
if(g[i]<minn){
ans=i;
minn=g[i];
}
}
printf("%lld\n",ans);
return 0;
}
string：
First flip, and then maintain the same color difference-set position
26 and then into k binary number, can not be determined for the point of using 26 binary numbers to determine k
Interval inversion balanced tree
We give 'a' to 'z' is a number [1, 26], to give each of '?' A different reference numeral 26 is greater than the input string into a string of numbers such, the count
of S.
We direct this string of numbers to do the operation, get a new string of numbers, denoted by T.
this thing is this classic problem bzoj3223:. Tyvj1729 literary balanced tree
directly to the S and T, and with every check and set up, the reference numeral represents characters represent necessarily the same.
If a link comprising at least one letter in the block, then the characters are identified by China Unicom block.
otherwise, there are 26 possible, can be directly assigned to sweep over.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<cmath>
#define int long long
using namespace std;
const int MAXN=5e5+5;
int n,m,k,l[MAXN],r[MAXN],b[MAXN];
char ch[MAXN];
bool pd[MAXN];
struct node1{
int val,pos;
}a[MAXN];
int fa[MAXN];
int find(int x){
return fa[x]=(fa[x]==x?x:find(fa[x]));
}
void merge(int x,int y){
x=find(x),y=find(y);
fa[max(x,y)]=min(x,y);
}
vector<int>v[MAXN];
int col[MAXN],tot=-1;
struct Node{
bool rev;
int s,h,v;
Node *l,*r;
}*root=0;
struct tbw{
Node *l,*r;
tbw(){}
tbw(Node *a,Node *b){l=a,r=b;}
};
int gets(Node *x){return x?x->s:0;}
void update(Node *x){x->s=gets(x->l)+1+gets(x->r);}
void rev(Node *x){
if(!x) return;
x->rev ^=1;
swap(x->l,x->r);
}
void down(Node *x){
if(x->rev) rev(x->l),rev(x->r),x->rev=0;
}
Node *merge(Node *a,Node *b){
if(!a) return b;
if(!b) return a;
if (a->h<b->h){
down(a);
a->r=merge(a->r,b);
update(a);
return a;
}
down(b);
b->l=merge(a,b->l);
update(b);
return b;
}
tbw split(Node *x,int k){
if(!x) return tbw(0,0);
down(x);
int tot=gets(x->l)+1;
if(k<tot){
tbw a=split(x->l,k);
x->l=a.r;
update(x);
return tbw(a.l,x);
}
tbw a=split(x->r,k-tot);
x->r=a.l;
update(x);
return tbw(x,a.r);
}
void insert(int v){
Node *x=new Node;
x->h=rand(),x->l=x->r=0;
x->v=v,x->s=1;
root=merge(root,x);
}
void rev(int l,int r){
tbw a=split(root,l-1);
tbw b=split(a.r,r-l+1);
rev(b.l);
root=merge(merge(a.l,b.l),b.r);
}
int num=0;
void print(Node *x){
if(!x) return;
down(x);
print(x->l);
b[++num]=x->v;
print(x->r);
}
signed main(){
scanf("%lld%lld%lld",&n,&m,&k);
scanf("%s",ch+1);
for(int i=1;i<=n;++i){
fa[i]=a[i].pos=b[i]=i;
a[i].val=0;
if(ch[i]!='?'){
a[i].val=ch[i]-'a'+1;
pd[i]=1;
}
insert(i);
}
for(int i=1;i<=m;++i){
scanf("%lld%lld",&l[i],&r[i]);
rev(l[i],r[i]);
}
print(root);
for(int i=1;i<=n;++i){
merge(b[i],i);
}
for(int i=1;i<=n;++i){
int x=find(i);
v[x].push_back(i);
if(a[x].val!=0) col[x]=a[x].val;
else if(a[i].val!=0) col[x]=a[i].val;
}
for(int i=1;i<=n;++i){
int x=find(i);
if(col[x]==0) continue;
a[i].val=col[x];
}
for(int i=1;i<=n;++i){
if(col[i]==0&&v[i].size()!=0) ++tot;
}
for(int i=1;i<=n;++i){
int x=find(i);
if(a[i].val!=0) continue;
else if(x!=i) a[i].val=a[x].val;
else if(tot>13) a[i].val=1,--tot;
else{
int p=pow(26,tot--);
int x=k/p;
if(k%p) ++x;
a[i].val=x--;
k-=x*p;
}
}
for(int i=1;i<=n;++i){
putchar(a[i].val+'a'-1);
}
puts("");
return 0;
}