Processing math: 100%

The Preliminary Contest for ICPC Asia Xuzhou 2019

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逐渐智障

B.so easy

并查集，可能会卡掉map,建议使用unordered_map。

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+100;
const int mod = 1e9+7;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll llINF = 0x3f3f3f3f3f3f3f3f;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define fep(i,a,b) for(int i=(a);i>=(b);i--)
inline bool read(ll &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    } 
    if(IsN) num=-num;
    return true;
}
unordered_map<int,int> par;
int find(int x){
    if(par[x]==0){
        par[x]=x;
        return x;
    }
    return x==par[x]?x:par[x]=find(par[x]);
}
int main(){
    int n,q;
    scanf("%d %d",&n,&q);
    int op,x;
    while(q--){
        scanf("%d %d",&op,&x);
        if(op==1){
            par[find(x)]=x+1;
        }else{
            if(find(x)>n) puts("-1");
            else printf("%d\n",find(x));
        }
    }
    return 0;
}

C.Buy Watermelon

签到，大于2的偶数都可以被拆分成两个偶数和。

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+100;
const int mod = 1e9+7;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll llINF = 0x3f3f3f3f3f3f3f3f;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define fep(i,a,b) for(int i=(a);i>=(b);i--)
inline bool read(ll &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    } 
    if(IsN) num=-num;
    return true;
}
int main(){
  //  freopen("1.in", "r", stdin);
    ll t,n;
    read(n);
    if(n>2&&(n%2==0)){
        cout<<"YES"<<endl;
    }
    else{
        cout<<"NO"<<endl;
    }
    return 0;
}

D. Carneginon

KMP,懒得写了（狗头）。。

E. XKC’s basketball team

对于每个i查找最右的大于等于ai+m的位置。线段树维护区间最大值优先查右子树即可。

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#include<bits/stdc++.h>
using namespace std;
const int N = 2e6+100;
const int mod = 1e9+7;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll llINF = 0x3f3f3f3f3f3f3f3f;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define fep(i,a,b) for(int i=(a);i>=(b);i--)
inline bool read(ll &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    } 
    if(IsN) num=-num;
    return true;
}
int mxx[N],a[N];
int n,m;
void pushup(int rt){
    mxx[rt]=max(mxx[rt<<1],mxx[rt<<1|1]);
}
void buildtree(int rt,int l,int r){
    mxx[rt]=0;
    if(l==r){
        mxx[rt]=a[l];return ;
    }
    int mid=(l+r)>>1;
    buildtree(rt<<1,l,mid);
    buildtree(rt<<1|1,mid+1,r);
    pushup(rt);
}
int query(int rt,int l,int r,int x){
   // cout<<rt<<' '<<l<<' '<<r<<' '<<mxx[rt]<<endl;
    if(l==r){
        if(mxx[rt]>=x){
             return l;
        }
        else return -1;
    }
    int mid=(l+r)>>1;
    if(mxx[rt<<1|1]>=x){
        return query(rt<<1|1,mid+1,r,x);//先查右面
    }else{
        return query(rt<<1,l,mid,x);
    }
}
int main(){
    //freopen("1.in", "r", stdin);
    scanf("%d %d",&n,&m);
    rep(i,1,n) scanf("%d",&a[i]);
    buildtree(1,1,n);
    vector<int> ans;
    rep(i,1,n){
        int id=query(1,1,n,a[i]+m);
        if(id<=i) ans.push_back(-1);
        else{
            ans.push_back(id-i-1);
        }
    }
    printf("%d",ans[0]);
    rep(i,1,(int)ans.size()-1) printf(" %d",ans[i]);
    puts("");
    return 0;
}

I. query

对于[1,n]，枚举它们所有的倍数，总对数也就大概只有O(n⋅log(n))，因此转化为一个正方形，每次查询[l,l]到[r,r]之间点的个数即可。接下来直接离线然后用树状数组维护前缀和即可。

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#include<bits/stdc++.h>
using namespace std;
const int N = 2e6+100;
const int mod = 1e9+7;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll llINF = 0x3f3f3f3f3f3f3f3f;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define fep(i,a,b) for(int i=(a);i>=(b);i--)
inline bool read(ll &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    } 
    if(IsN) num=-num;
    return true;
}

int n,m;
int c[N];
int lowbit(ll x){
    return (x&(-x));
}
void add(ll x,ll v){
    for(;x<N;x+=lowbit(x)){
         c[x]+=v;
         //cout<<x<<' '<<v<<endl;
    }
}
ll query(ll x){
    ll ans=0;
    for(;x;x-=lowbit(x)){
        ans+=c[x];
    }
    return ans;
}

int pos[N],a[N];
int l[N],r[N];
struct point{
    int x,y,flag;
}s[N];
bool cmp(point a,point b){
    if(a.x==b.x){
        if(a.y==b.y){
           return a.flag<b.flag;
        }
        return a.y<b.y;
    }
    return a.x<b.x;
}
int yy[N],id[N];
map<pair<int,int>,int> mmp;

int main(){
    scanf("%d %d",&n,&m);
    rep(i,1,n){
         scanf("%d",&a[i]);
         pos[a[i]]=i;
    }
    int cnt=0;
    rep(i,1,n){
        for(int j=2*i;j<=n;j+=i){
            s[++cnt]={pos[i],pos[j],0};
        }
    }
    rep(i,1,m){
        scanf("%d %d",&l[i],&r[i]);
        s[++cnt]={r[i],r[i],1};
        s[++cnt]={l[i]-1,r[i],1};
        s[++cnt]={r[i],l[i]-1,1};
        s[++cnt]={l[i]-1,l[i]-1,1};
    }

    rep(i,1,cnt) yy[i]=s[i].y;

    sort(yy+1,yy+cnt+1);
    int siz=unique(yy+1,yy+cnt+1)-yy-1;

    sort(s+1,s+cnt+1,cmp);
    rep(i,1,cnt){
        id[i]=lower_bound(yy+1,yy+siz+1,s[i].y)-yy;
    }
    rep(i,1,cnt){
        if(s[i].flag==0){
            add(id[i],1);
        }else{
            mmp[{s[i].x,s[i].y}]=query(id[i]);
        }
    }
    rep(i,1,m){
        printf("%d\n",mmp[{r[i],r[i]}]-mmp[{l[i]-1,r[i]}]-mmp[{r[i],l[i]-1}]+mmp[{l[i]-1,l[i]-1}]);
    }
    return 0;
}

J. Random Access Iterator

首先预处理出每个点所能到达的最大深度和子节点的数量，令dp[x]表示能到达叶子节点的概率，当从u向v转移时，只有v能到达的的最大深度等于树的深度时才能转移。对于叶子节点：dp[x]=1,向上转移即可。

dp[u]=1−(1−∑dp[v]|siz[u]|)|siz[u]|
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#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+100;
const int mod = 1e9+7;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll llINF = 0x3f3f3f3f3f3f3f3f;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define fep(i,a,b) for(int i=(a);i>=(b);i--)
inline bool read(ll &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    } 
    if(IsN) num=-num;
    return true;
}
ll qpow(ll a,ll b){
    ll ans=1;
    while(b){
        if(b&1) ans=(ans%mod*a%mod)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ans%mod;
}
ll inv(ll a){
    return qpow(a,mod-2);
}
vector<int> G[N];
int depth[N],maxxh[N],siz[N];
void dfs(int u,int par){
    depth[u]=depth[par]+1;
    maxxh[u]=depth[u];
    for(int v:G[u]){
        if(v==par) continue;
        siz[u]++;
        dfs(v,u);
        maxxh[u]=max(maxxh[u],maxxh[v]);
    }
}
ll dp[N];
ll solve(ll u,ll fa){
    if(siz[u]==0) return dp[u]=1;
    ll ans=0;
    for(int v:G[u]){
        if(v==fa) continue;
        if(maxxh[v]==maxxh[1]){
            ans+=solve(v,u);
        }
    }
    ans=ans*inv(siz[u])%mod;
    return dp[u]=(1-qpow(1-ans+mod,siz[u])+mod)%mod;
}
int main(){
    int n,u,v;
    cin>>n;
    rep(i,1,n-1){
        cin>>u>>v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    dfs(1,1);
    cout<<solve(1,1)<<endl;
    return 0;
}

M. Longest subsequence

在S中找到最长的子序列使其字典序大于T,输出长度即可。

假设子序列为T′，如果T′>T,则T′一定存在某个位置p使得T′[1…p−1]=T[1…p]，而T′[p]>T[p],枚举p，利用序列自动机计枚举T′[p]并快速求的位置，计算答案即可。

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#include<bits/stdc++.h>
using namespace std;
const int N = 1e6+100;
const int mod = 1e9+7;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll llINF = 0x3f3f3f3f3f3f3f3f;
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define fep(i,a,b) for(int i=(a);i>=(b);i--)
inline bool read(ll &num) {
    char in;bool IsN=false;
    in=getchar();
    if(in==EOF) return false;
    while(in!='-'&&(in<'0'||in>'9')) in=getchar();
    if(in=='-'){ IsN=true;num=0;}
    else num=in-'0';
    while(in=getchar(),in>='0'&&in<='9'){
            num*=10,num+=in-'0';
    } 
    if(IsN) num=-num;
    return true;
}
int nxt[N][27];
void init(char *s){
    int l=strlen(s);
    for(int i=0;i<26;i++) nxt[l][i]=INF;
    for(int i=l-1;i>=0;i--){
        for(int j=0;j<26;j++){
            nxt[i][j]=nxt[i+1][j];
        }
        nxt[i][s[i]-'a']=i;
    }
}
char s[N],t[N];
int main(){
    //freopen("1.in", "r", stdin);
    int n,m;
    cin>>n>>m;
    scanf("%s %s",s,t);
    init(s);
    int pos=-1;
    int ans=0;
    int cnt=0;
    bool flag=0;
    vector<int> id;
    for(int i=0;i<n;i++){
        for(int j=t[cnt]-'a'+1;j<26;j++){
            if(nxt[i][j]!=INF){
                flag=1;
                ans=max(ans,cnt+n-nxt[i][j]);
            }
        }
        if(s[i]==t[cnt]){
            cnt++;
            id.push_back(i);
        }
        if(cnt==m) break;
    }
    if(!flag){
        if(id.size()==m){
            if(id[m-1]==n-1)cout<<-1<<endl;
            else cout<<m+(n-id[m-1]-1)<<endl;
        }
        else cout<<-1<<endl;
    }
    else{
        if(id.size()==m&&id[m-1]!=-1){
            ans=max(ans,m+(n-id[m-1]-1));
        }
        cout<<ans<<endl;
    }
    return 0;
}