The 2018 ACM-ICPC Asia Nanjing Regional Programming Contest

A. Adrien and Austin

大意: $n$个石子, 编号$1$到$n$, 两人轮流操作, 每次删除$1$到$k$个编号连续的石子, 不能操作则输, 求最后胜负情况.

删除一段后变成两堆, 可以用$sg$函数打表找规律

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
    int n,k;
    cin>>n>>k;
    if (!n) return puts("Austin"),;
    if (k!=) return puts("Adrien"),;
    return puts(n&?"Adrien":"Austin");
}

D. Country Meow

大意: 给定空间$n$个点, 求构造一个点到$n$个点距离和最小.

等价于求一个最小的球覆盖所有的点, 可以三分套三分

E. Eva and Euro coins

大意: 给定$01$串$s$和$t$, 每次操作在$s$中选择一段连续相同的长为$k$的区间翻转, 求$s$是否能变为$t$

观察$1$: 操作可逆. 观察$2$: 长度为$k$的连续相同区间可以任意移动.

注意特判$k=1$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head

const int N = 1e6+;
int n,k;
char s[N],t[N];
pii a[N];

void trans(char s[], int n) {
    int top = ;
    REP(i,,n) {
        if (!top||a[top].x!=s[i]) a[++top]=pii(s[i],);
        else if (++a[top].y==k) --top;
    }
    int cnt = ;
    REP(i,,top) while (a[i].y--) s[++cnt] = a[i].x;
    REP(i,cnt+,n) s[i] = '';
}

int main() {
    scanf("%d%d%s%s",&n,&k,s+,t+);
    if (k==) return puts("Yes"),;
    trans(s,n),trans(t,n);
    puts(strcmp(s+,t+)?"No":"Yes");
}

G. Pyramid

大意: 求$n$层等边三角形中等边三角形的个数.

打表然后插值求系数, 答案为$\frac{n}{4}+\frac{11n^2}{24}+\frac{n^3}{4}+\frac{n^4}{24}$

#include <iostream>
#include <set>
#include <queue>
#define x first
#define y second
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;
typedef pair<int,int> pii;
int n;
set<pii> s;

void dfs(int d, int x, int y) {
    if (d>n) return;
    if (s.count(pii(x,y))) return;
    s.insert(pii(x,y));
    dfs(d+,x-,y+);
    dfs(d+,x+,y+);
}
int dis(pii u, pii v) {
    return (u.x-v.x)*(u.x-v.x)+*(u.y-v.y)*(u.y-v.y);
}
int main() {
    cin>>n;
    dfs(,,);
    vector<pii> v(s.begin(),s.end());
    int sz = v.size(), ans = ;
    REP(i,,sz-) REP(j,i+,sz-) REP(k,j+,sz-) {
        int q=dis(v[i],v[j]),w=dis(v[i],v[k]),e=dis(v[j],v[k]);
        if (q==w&&w==e) ++ans;
    }
    printf("%d\n",ans);
}

暴力代码

H. Huge Discount

大意: 给定一个长$n$的串$s$, 只含数字'0','1','2', 有$n$个商品, 第$i$个商品价格为s[i...n]. 对于一个商品, 每次操作选两个连续且不相同的数删除, 可以进行任意次操作. 求$n$个商品最小价格和.

I. Magic Potion

大意: $n$个英雄, $m$个怪物, 给定每个英雄能杀死的怪物集合, 每个英雄只能从中选一个杀死. 可以选出$k$个英雄多杀一次, 求最大能杀怪物数.

最大流水题, 开个虚拟结点, 源点向虚拟结点连$k$即可.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
const int N = 1e6+;
const int SS = N-, S = N-, T = N-, INF = 0x3f3f3f3f;
int n, m, k;
struct edge {
    int to,w,next;
    edge(int to=,int w=,int next=):to(to),w(w),next(next){}
} e[N];
int head[N], dep[N], vis[N], cur[N], cnt=;
queue<int> Q;
int bfs() {
    REP(i,,n+m) dep[i]=INF,vis[i]=,cur[i]=head[i];
    REP(i,SS,T) dep[i]=INF,vis[i]=,cur[i]=head[i];
    dep[S]=,Q.push(S);
    while (Q.size()) {
        int u = Q.front(); Q.pop();
        for (int i=head[u]; i; i=e[i].next) {
            if (dep[e[i].to]>dep[u]+&&e[i].w) {
                dep[e[i].to]=dep[u]+;
                Q.push(e[i].to);
            }
        }
    }
    return dep[T]!=INF;
}
int dfs(int x, int w) {
    if (x==T) return w;
    int used = ;
    for (int i=cur[x]; i; i=e[i].next) {
        cur[x] = i;
        if (dep[e[i].to]==dep[x]+&&e[i].w) {
            int f = dfs(e[i].to,min(w-used,e[i].w));
            if (f) used+=f,e[i].w-=f,e[i^].w+=f;
            if (used==w) break;
        }
    }
    return used;
}
int dinic() {
    int ans = ;
    while (bfs()) ans+=dfs(S,INF);
    return ans;
}
void add(int u, int v, int w) {
    e[++cnt] = edge(v,w,head[u]);
    head[u] = cnt;
    e[++cnt] = edge(u,,head[v]);
    head[v] = cnt;
} 

int main() {
    scanf("%d%d%d",&n,&m,&k);
    add(S,SS,k);
    REP(i,,n) {
        add(S,i,);
        add(SS,i,);
        int t;
        scanf("%d",&t);
        while (t--) {
            int x;
            scanf("%d",&x);
            add(i,n+x,);
        }
    }
    REP(i,,m) add(n+i,T,);
    printf("%d\n",dinic());
}

J. Prime Game

大意: 给定序列$a$, 设$fac(l,r)$为$[l,r]$区间积的素因子个数, 求$\sum\limits_{i=1}^n\sum\limits_{j=i}^n fac(i,j)$

计算出每种素因子所在的区间即可

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, P2 = , INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head

const int N = 1e6+;
int n, mi[N], a[N];
vector<int> g[N];
int main() {
    REP(i,,N-) mi[i]=i;
    REP(i,,N-) if (mi[i]==i) {
        for (int j=i;j<N;j+=i) mi[j]=min(mi[j],i);
    }
    scanf("%d", &n);
    REP(i,,n) {
        scanf("%d", a+i);
        while (a[i]!=) g[mi[a[i]]].pb(i),a[i]/=mi[a[i]];
    }
    ll ans = ;
    REP(i,,N-) if (g[i].size()) {
        ll t = (ll)n*(n+)/;
        int now = ;
        g[i].pb(n+);
        for (int j:g[i]) {
            t -= (ll)(j-now)*(j-now+)/;
            now = j+;
        }
        ans += t;
    }
    printf("%lld\n", ans);
}

M. Mediocre String Problem

大意: 给定串$s,t$, 求三元组$(i,j,k)$的个数, 满足$j-i+1>k$且$s[i,j]+t[1,k]$为回文串

合法方案一定是$s$中与$t$的一段前缀对称, 然后中间再连上一段回文串.

exkmp求出最长匹配前缀, pam求出每个位置开头的回文串数

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head

const int N = 4e6+;
void init(char *s, int *z, int n) {
    int mx=,l=;
    REP(i,,n-) {
        z[i] = i<mx?min(mx-i,z[i-l]):;
        while (s[z[i]]==s[i+z[i]]) ++z[i];
        if (i+z[i]>mx) mx=i+z[i],l=i;
    }
}

int n, m, tot, cnt, last, fac[N];
int fail[N], len[N], ch[N][], num[N];
ll sum[N];
char ss[N];

int getfail(int x) {
    while (ss[cnt-len[x]-]!=ss[cnt]) x=fail[x];
    return x;
}
int insert(int c) {
    ss[++cnt] = c;
    int p = getfail(last);
    if (!ch[p][c]) {
        len[++tot] = len[p]+;
        fail[tot]=ch[getfail(fail[p])][c];
        ch[p][c]=tot;
        num[tot]=num[fail[tot]]+;
    }
    last = ch[p][c];
    return num[last];
}
void clear() {
    REP(i,,tot) {
        memset(ch[i],,sizeof ch[]);
        num[i]=len[i]=fail[i]=;
    }
    ss[]='#',fail[]=,last=;
    len[]=,len[]=-,tot=,cnt=;
}

char s[N], t[N];
int c[N],z[N];

int main() {
    scanf("%s%s",s,t);
    n = strlen(s);
    m = strlen(t);
    clear();
    PER(i,,n-) c[i]=insert(s[i]);
    reverse(s,s+n);
    t[m] = '?';
    strcat(t,s);
    init(t,z,n+m+);
    ll ans = ;
    REP(i,m+,n+m) ans += (ll)z[i]*c[n+-i+m];
    printf("%lld\n", ans);
}