\(A\)
Gugu Gu
const int N=1e6+5;
char s[N],t[N];int n,res;
inline bool cmp(const int &x,const int &y){return x>y;}
int main(){
scanf("%s",s+1),n=strlen(s+1);
fp(i,1,n)t[i]=s[i];sort(t+1,t+1+n,cmp);
fp(i,1,n)res=(res*10+t[i]-s[i]),res=(res%10+10)%10;
printf("%d\n",res);
return 0;
}\(B\)
A first pretreatment \ (dp [i] \) represents the cost \ (I \) cost up to the \ (n-\) returns obtained after day, then directly \ (DP \) on the line
//quming
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
typedef long long ll;
const int N=10005;
ll dp[N],f[2][N],res;int w[N],n,m,k,t;
inline int calc(R int x){return x<=k?w[x]:0;}
int main(){
scanf("%d%d%d",&n,&m,&k);
fp(i,0,k)scanf("%d",&w[i]);
memset(dp,0xef,sizeof(dp)),dp[0]=0;
for(R int a,b;m;--m){
scanf("%d%d",&a,&b);
fp(i,a,2000)if(dp[i-a]>=0)cmax(dp[i],dp[i-a]+b);
}
memset(f[0],0xef,sizeof(f[0])),f[t=0][calc(0)]=0;
fp(i,1,n-1){
memset(f[t^1],0xef,sizeof(f[t^1]));
fp(j,0,2000)if(f[t][j]>=0){
// printf("%d %d %lld\n",i,j,f[t][j]);
fp(l,0,j)
cmax(f[t^1][j-l+calc(j-l)],f[t][j]+dp[l]);
}
t^=1;
}
res=-1e18;
fp(i,0,2000)if(f[t][i]>=0)cmax(res,f[t][i]+i);
printf("%lld\n",res);
return 0;
}\(C\)
I think there should be only one person I was \ (zz \) to write a \ (O (n \ log n ) \) is the result of a large constant practice actually ran past people ......
A practice
First, obviously can not exceed the length of the segment \ ({(n--. 1) \ over 2} \) . Breaking the ring into the first strand, we remember \ (las_i \) represented by \ (I \) ending minimum \ (J \) satisfies \ ([j + 1, i ] \) is a valid interval, empathy defined \ (nxt_i \) represents the maximum \ (J \) satisfies \ ([i, j-1 ] \) is a is a valid interval
We enumerate the end of a range in which it finally got the answer divided by \ (3 \) is the original answer
Suppose the end of \ (T \) , then the beginning of the loop is \ (S = T-n-+. 1 \) , we need to number that satisfies \ (i \ in [las_j, j-1], i \ in [ s, nxt_s-1], j \ in [las_t, i-1] \) of the number of \ (i, j \) number
We can each \ (J \) seen for a \ ([las_j, j-1 ] \) interval plus, with pointers to ensure that only \ (j \ in [las_t, i-1] \) section contribute to, and then queries the interval \ ([s, nxt_s-1 ] \) on it. Starting with a segment tree seems \ (MLE \) , and later changed to Fenwick tree cards for a long time often results last \ (955ms \) over this question ......
Practice two
Into a chain that does not damage the definition is slightly altered, denoted \ (las_i \) represented by \ (i-1 \) ending minimum \ (J \) satisfies \ ([j, i] \ ) is a legitimate the interval (only considering \ (j> i \) of those, that this range is actually \ ([J, the n-] | [1, i-1] \) ), and \ (nxt_i \) indicates the maximum the \ (j \) satisfies \ ([i, j] \ ) is a legitimate range
We enumerate \ (A, b \) , you will need to meet the condition that \ (b \ in [a + 1, nxt_a + 1], c \ in [b + 1, nxt_b + 1], c \ in [las_a , n] \)
For convenience, we put all \ (nxt_b> n-1 \ ) are set to \ (the n--1 \) , can be seen not affect the answer
Because the length of each section is less than equal to \ ({(n--. 1) \ over 2} \) , so \ (b <las_a \) constant established, then the \ (C \) is the value of the program is legitimate \ ( nxt_b + 2-las_a \)
Then we enumerate \ (A \) , with a two-pointer maintain all legal \ (b \) contribution to
Specific details can reference code
Code a
//quming
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
typedef long long ll;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
R int res,f=1;R char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
const int N=4e6+5;
ll d[N];int c[N];
int a[N],las[N],nxt[N],pos[N],n,m,len;ll res;
inline int max(R int x,R int y){return x>y?x:y;}
inline int min(R int x,R int y){return x<y?x:y;}
inline void chgd(R int x,R int v){for(;x<=m;x+=x&-x)d[x]+=v;}
inline void chgc(R int x,R int v){for(;x<=m;x+=x&-x)c[x]+=v;}
inline void upd(R int l,R int r,R int v){
chgc(r+1,-v),chgd(r+1,-v*(r+1));
if(l!=0)chgc(l,v),chgd(l,v*l);
}
inline int askc(R int x){R int res=0;for(;x;x-=x&-x)res+=c[x];return res;}
inline ll askd(R int x){R ll res=0;for(;x;x-=x&-x)res+=d[x];return res;}
inline ll query(R int l,R int r){
R ll res=0;
res+=1ll*(r+1)*askc(r)-askd(r);
if(l>1)res-=1ll*l*askc(l-1)-askd(l-1);
return res;
}
int main(){
// freopen("testdata.in","r",stdin);
n=read(),m=n<<1;
fp(i,1,n)a[i]=read(),a[n+i]=a[i];
len=(n-1)>>1;
las[0]=0,nxt[n<<1|1]=(n<<1|1);
fp(i,1,n<<1)las[i]=max(las[i-1],max(i-len,pos[a[i]])),pos[a[i]]=i;
fp(i,1,n)pos[i]=(n<<1|1);
fd(i,n<<1,1)nxt[i]=min(nxt[i+1],min(i+len,pos[a[i]])),pos[a[i]]=i;
// fp(i,2,n)upd(las[i],i-1,1);
fp(i,las[n+1],n)upd(las[i],i-1,1);
for(R int i=n+1,j=las[n+1];i<=(n<<1);++i){
while(j<las[i])upd(las[j],j-1,-1),++j;
res+=query(i-n+1,nxt[i-n+1]-1);
upd(las[i],i-1,1);
}
printf("%lld\n",res/3);
return 0;
}Code two
//minamoto
#include<bits/stdc++.h>
#define R register
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
typedef long long ll;
char buf[1<<21],*p1=buf,*p2=buf;
inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
int read(){
R int res,f=1;R char ch;
while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
return res*f;
}
const int N=4e6+5;
int a[N],las[N],nxt[N],pos[N],c[N],n,m,len,cnt;ll res,sum;
inline int max(R int x,R int y){return x>y?x:y;}
inline int min(R int x,R int y){return x<y?x:y;}
int main(){
// freopen("testdata.in","r",stdin);
n=read(),len=(n-1)>>1;
fp(i,1,n)a[i]=read(),a[n+i]=a[i];
las[0]=0,nxt[n<<1|1]=(n<<1|1);
fp(i,1,n<<1)las[i]=max(las[i-1],max(i-len+1,pos[a[i]]+1)),pos[a[i]]=i;
fp(i,1,n)pos[i]=(n<<1|1);
fd(i,n<<1,1)nxt[i]=min(nxt[i+1],min(i+len-1,pos[a[i]]-1)),pos[a[i]]=i;
fp(i,1,n)las[i]=las[i+n-1],cmin(nxt[i],n-1);
/*
fp(i,1,n)if(las[i]>i){
fp(j,i+1,nxt[i]+1)if(nxt[j]+1>=las[i])res+=nxt[j]+2-las[i];
}else break;
*/
for(R int i=1,j=1,l=1,r=1;i<=n;++i){
if(las[i]<=i)break;
while(r<=nxt[i]+1)++cnt,++c[nxt[r]+1],sum+=nxt[r]+2,++r;
while(l<r&&(l<=i||nxt[l]+1<las[i]))--cnt,--c[nxt[l]+1],sum-=nxt[l]+2,++l;
res+=sum-1ll*cnt*las[i];
}
printf("%lld\n",res);
return 0;
}\(D\)
Similar to \ (kruskal \) the process of reconstruction of the tree, we enumerate point from small to large, and only consider those side from Dalian to small. For new out of this tree, we want to ensure that each node point less than his father, so the current point to enumerate all the edges, the communication block corresponding to a sub-tree of the current point on the line
Then this is similar to \ (kruskal \) what reconstruction of a tree, asking when multiplied \ (+ \) tree line on the line
//quming
#include<bits/stdc++.h>
#define R register
#define pb emplace_back
#define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
#define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,1:0;}
using namespace std;
const int P=998244353;
inline void upd(R int &x,R int y){(x+=y)>=P?x-=P:0;}
inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}
inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}
inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}
int ksm(R int x,R int y){
R int res=1;
for(;y;y>>=1,x=mul(x,x))(y&1)?res=mul(res,x):0;
return res;
}
const int N=5e5+5;
vector<int>to[N];
struct eg{int v,nx;}e[N<<1];int head[N],tot;
inline void Add(R int u,R int v){e[++tot]={v,head[u]},head[u]=tot;}
int a[N],rk[N],dfn[N],low[N],fa[N][35],ga[N],dep[N],Log[N],tim,n,m,q;
inline int find(R int x){return ga[x]==x?x:ga[x]=find(ga[x]);}
struct node;typedef node* ptr;
struct node{
ptr lc,rc;int v;
inline void upd(){v=mul(lc->v,rc->v);}
}ee[N<<2],*pp=ee,*rt;
void build(ptr &p,int l,int r){
p=++pp;if(l==r)return p->v=a[rk[r]],void();
int mid=(l+r)>>1;
build(p->lc,l,mid),build(p->rc,mid+1,r);
p->upd();
}
void update(ptr p,int l,int r,int x,int v){
if(l==r)return p->v=v,void();
int mid=(l+r)>>1;
x<=mid?update(p->lc,l,mid,x,v):update(p->rc,mid+1,r,x,v);
p->upd();
}
int query(ptr p,int l,int r,int ql,int qr){
if(ql<=l&&qr>=r)return p->v;
int mid=(l+r)>>1,res=1;
if(ql<=mid)res=mul(res,query(p->lc,l,mid,ql,qr));
if(qr>mid)res=mul(res,query(p->rc,mid+1,r,ql,qr));
return res;
}
void dfs(int u,int f){
dfn[u]=++tim,rk[tim]=u,fa[u][0]=f;
fp(i,1,20)fa[u][i]=fa[fa[u][i-1]][i-1];
go(u)dfs(v,u);
low[u]=tim;
}
inline int jump(R int u,R int x){
if(u>x)return 0;
// R int k=dep[u];
fd(i,20,0)if(fa[u][i]&&fa[u][i]<=x)u=fa[u][i];
return query(rt,1,n,dfn[u],low[u]);
}
int main(){
// freopen("testdata.in","r",stdin);
scanf("%d%d%d",&n,&m,&q);
fp(i,1,n)scanf("%d",&a[i]),ga[i]=i,a[i]%=P;
for(R int i=1,u,v;i<=m;++i)scanf("%d%d",&u,&v),to[u].pb(v),to[v].pb(u);
fp(u,1,n)for(auto v:to[u])if(v<u&&u!=find(v))Add(u,find(v)),ga[find(v)]=u;
// fp(i,1,n)fa[i][0]=ga[i];
Log[0]=-1;fp(i,2,n)Log[i]=Log[i>>1]+1;
dfs(n,0);
build(rt,1,n);
for(int op,u,v;q;--q){
scanf("%d%d%d",&op,&u,&v);
if(op==1)printf("%d\n",jump(u,v));
else update(rt,1,n,dfn[u],v%P);
}
return 0;
}The remaining two questions tomorrow to fix it