2017 ACM Arabella Collegiate Programming Contest(solved 9/13, complex 12/13)

时间：2021-11-12 07:23:54

A.Sherlock Bones

题意: 给出长度为n的01串，问f(i,j)=f(j,k),(i<j<k)的i,j,k取值种数。其中f(i,j)表示[i,j]内1的个数， 且s[j]必须为1。

先把串看出是一串1两两之间穿插若干个0的联通块，不妨设block[i]为联通块i里面0的个数。

先考虑i，k处都为0的情况。

枚举i在哪个联通块里面。再枚举j，由于对称性，此时k在block[i+1],block[i+3],block[i+5]。。。内，那么此时方法数为block[i]*(block[i+1]+block[i+3]+...)，前缀和预处理优化一下即可。

i，k处其中至少一个为1的情况同理。

时间复杂度O(n).

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <bitset>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FDR(i,a,n) for(int i=a; i>=n; --i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

inline int Scan() {

    int x=,f=; char ch=getchar();

    while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}

    return x*f;

}

inline void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

char s[N];

int num[N], sum1[N], sum2[N];

int main ()

{

    int T, n;

    scanf("%d",&T);

    while (T--) {

        scanf("%d%s",&n,s);

        int pos=;

        LL ans=;

        mem(num,);

        FOR(i,,n-) {

            if (s[i]=='') ++pos;

            else ++num[pos];

        }

        for (int i=; i<=pos; i+=) sum1[i]=num[i]+(i>=?sum1[i-]:);

        for (int i=; i<=pos; i+=) sum2[i]=num[i]+sum2[i-];

        int top1=(pos&?pos:pos-), top2=(pos&?pos-:pos);

        FOR(i,,pos-) {

            if (i&) {

                ans+=(LL)num[i]*(sum2[i+]-sum2[i-]);

                ans+=(LL)(num[i]+)*(sum2[top2]-sum2[i+]+(top2-i-)/);

            }

            else {

                ans+=(LL)num[i]*(sum1[i+]-sum1[i-]);

                ans+=(LL)(num[i]+)*(sum1[top1]-sum1[i+]+(top1-i-)/);

            }

        }

        printf("%lld\n",ans);

    }

    return ;

}

B.Unusual Team

水题。

C.Cheap Kangaroo

水题。

D.Magical Bamboos

题意:给出n个数，可以给其中某个数减1，其他的数加1，问是否可能最后全相等。

操作实际上就是给其中某个数减2，因此判断他们的奇偶性即可。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <bitset>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FDR(i,a,n) for(int i=a; i>=n; --i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

inline int Scan() {

    int x=,f=; char ch=getchar();

    while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}

    return x*f;

}

inline void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

int a[N];

int main ()

{

    int T=Scan();

    while (T--) {

        int n=Scan();

        FOR(i,,n) a[i]=Scan();

        int tmp=a[]&, flag=;

        FOR(i,,n) if (a[i]%!=tmp) {flag=; break;}

        puts(flag?"yes":"no");

    }

    return ;

}

E.Competitive Seagulls

题意:给出一串长度为n的白棋子，两位操作者轮流将p<=ceil(n/2)的连续棋子涂黑，其中p为质数，若不存在这样的p，则p=1，无法操作者输，问谁能赢？

结论:n=2或者n=3，则第二位赢，否则第一位赢。

可以构造出来，当n!=2并且n!=3时，如果n为奇数，第一位涂最中间的3颗棋子，之后无论第二位怎么涂，只需涂第二位涂的对称位置即可。

如果n为偶数，第一位涂最中间的2颗棋子，其余同上。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <bitset>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FDR(i,a,n) for(int i=a; i>=n; --i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

inline int Scan() {

    int x=,f=; char ch=getchar();

    while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}

    return x*f;

}

inline void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

int main ()

{

    int T=Scan();

    while (T--) {

        int x=Scan();

        puts(x!=&&x!=?"first":"second");

    }

    return ;

}

F.Monkeying Around

题意是有N个猴子在座位上讲笑话，每个猴子讲的笑话会有种类和范围，当一个猴子听到没有听过的笑话的时候，它会从座位上离开，掉地上；当听到已经听过的笑话的时候，它会从地上回到座位。现在给出各个讲笑话的情况，问最终猴子有多少在座位上。

把这个转换成区间问题，按照所有区间的左端点排序，右端点排序。枚举每一个猴子，维护当前影响该点的笑话，如果最后的笑话出现2次以上，那么该猴子在座位上。

#include <bits/stdc++.h>

#define IOS ios::sync_with_stdio(false)

using namespace std;

#define inf (0x3f3f3f3f)

typedef long long int LL;

const int maxn = 2e5 + ;

struct Node {

    int L, R, id;

}one[maxn], two[maxn];

bool cmp1(struct Node a, struct Node b) {

    return a.L < b.L;

}

bool cmp2(struct Node a, struct Node b) {

    return a.R < b.R;

}

int idForJoke[maxn];

int has[maxn];

set<int>ss;

void work() {

    ss.clear();

    memset(has, , sizeof has);

    int n, m;

    scanf("%d%d", &n, &m);

    for (int i = ; i <= m; ++i) {

        int pos, joke, dis;

        scanf("%d%d%d", &pos, &joke, &dis);

        one[i].L = max(, pos - dis), one[i].R = min(n, pos + dis), one[i].id = i;

        two[i] = one[i];

        idForJoke[i] = joke;

    }

    sort(one + , one +  + m, cmp1);

    sort(two + , two +  + m, cmp2);

    int ans = , p1 = , p2 = ;

    for (int i = ; i <= n; ++i) {

        while (p1 <= m && i >= one[p1].L) {

            ss.insert(one[p1].id);

            has[idForJoke[one[p1].id]]++;

            ++p1;

        }

        while (p2 <= m && i > two[p2].R) {

            ss.erase(two[p2].id);

            has[idForJoke[two[p2].id]]--;

            ++p2;

        }

        if (ss.size()) {

            ans += has[idForJoke[*ss.rbegin()]] > ;

        } else ans++;

    }

    printf("%d\n", ans);

}

int main() {

#ifdef local

    freopen("data.txt", "r", stdin);

//    freopen("data.txt", "w", stdout);

#endif

    int t;

    scanf("%d", &t);

    while (t--) work();

    return ;

}

G.Snake Rana

2017 ACM Arabella Collegiate Programming Contest(solved 9/13, complex 12/13)

H.Mirrored String I

水题。

I.Mirrored String II

题意：求字符串出现给定字符的最长回文子串。

非给定字符在原字符串中分割出一些子串，对这些子串使用马拉车算法更新答案即可。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <bitset>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi acos(-1.0)

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FDR(i,a,n) for(int i=a; i>=n; --i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

inline int Scan() {

    int x=,f=; char ch=getchar();

    while(ch<''||ch>''){if(ch=='-') f=-; ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-''; ch=getchar();}

    return x*f;

}

inline void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

char Ma[N<<];

int Mp[N<<];

char a[]={'A', 'H', 'I', 'M', 'O', 'T', 'U', 'V','W','X','Y'};

void Manacher(char *s, int len){

    int l=;

    Ma[l++]='$'; Ma[l++]='#';

    FOR(i,,len-) Ma[l++]=s[i], Ma[l++]='#';

    Ma[l]=;

    int mx=, id=;

    FOR(i,,l-) {

        Mp[i]=mx>i?min(Mp[*id-i],mx-i):;

        while (Ma[i+Mp[i]]==Ma[i-Mp[i]]) Mp[i]++;

        if (i+Mp[i]>mx) mx=i+Mp[i], id=i;

    }

}

char s[N];

int main ()

{

    int T=Scan();

    while (T--) {

        scanf("%s",s);

        int len=strlen(s);

        int now=, ans=;

        FOR(i,,len-) {

            bool mark=false;

            FOR(j,,) if (s[i]==a[j]) {mark=true; break;}

            if (!mark) {

                if (now<=i-) {

                    Manacher(s+now,i-now);

                    FOR(k,,*(i-now)+) ans=max(ans,Mp[k]-);

                }

                now=i+;

            }

        }

        if (now<=len-) {

            Manacher(s+now,len-now);

            FOR(k,,*(len-now)+) ans=max(ans,Mp[k]-);

        }

        printf("%d\n",ans);

    }

    return ;

}

J.Lazy Physics Cat

数学水题。

K.Owl Geeks

数学水题。

M.Make Cents?

水题。

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