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

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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1; char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}
inline void Out(int a) {if(a<0) {putchar('-'); a=-a;}if(a>=10) Out(a/10);putchar(a%10+'0');
}
const int N=200005;
//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=1;LL ans=0;mem(num,0);FOR(i,0,n-1) {if (s[i]=='1') ++pos;else ++num[pos];}for (int i=1; i<=pos; i+=2) sum1[i]=num[i]+(i>=2?sum1[i-2]:0);for (int i=2; i<=pos; i+=2) sum2[i]=num[i]+sum2[i-2];int top1=(pos&1?pos:pos-1), top2=(pos&1?pos-1:pos);FOR(i,1,pos-1) {if (i&1) {ans+=(LL)num[i]*(sum2[i+1]-sum2[i-1]);ans+=(LL)(num[i]+1)*(sum2[top2]-sum2[i+1]+(top2-i-1)/2);}else {ans+=(LL)num[i]*(sum1[i+1]-sum1[i-1]);ans+=(LL)(num[i]+1)*(sum1[top1]-sum1[i+1]+(top1-i-1)/2);}}printf("%lld\n",ans);}return 0;
}

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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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1; char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}
inline void Out(int a) {if(a<0) {putchar('-'); a=-a;}if(a>=10) Out(a/10);putchar(a%10+'0');
}
const int N=100005;
//Code begin...int a[N];int main ()
{int T=Scan();while (T--) {int n=Scan();FOR(i,1,n) a[i]=Scan();int tmp=a[1]&1, flag=1;FOR(i,2,n) if (a[i]%2!=tmp) {flag=0; break;}puts(flag?"yes":"no");}return 0;
}

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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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1; char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}
inline void Out(int a) {if(a<0) {putchar('-'); a=-a;}if(a>=10) Out(a/10);putchar(a%10+'0');
}
const int N=1005;
//Code begin...int main ()
{int T=Scan();while (T--) {int x=Scan();puts(x!=2&&x!=3?"first":"second");}return 0;
}

View Code

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 + 20;
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, 0, sizeof has);int n, m;scanf("%d%d", &n, &m);for (int i = 1; i <= m; ++i) {int pos, joke, dis;scanf("%d%d%d", &pos, &joke, &dis);one[i].L = max(1, pos - dis), one[i].R = min(n, pos + dis), one[i].id = i;two[i] = one[i];idForJoke[i] = joke;}sort(one + 1, one + 1 + m, cmp1);sort(two + 1, two + 1 + m, cmp2);int ans = 0, p1 = 1, p2 = 1;for (int i = 1; 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()]] > 1;} else ans++;}printf("%d\n", ans);
}int main() {
#ifdef localfreopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endifint t;scanf("%d", &t);while (t--) work();return 0;
}

View Code

G.Snake Rana

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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0,f=1; char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;
}
inline void Out(int a) {if(a<0) {putchar('-'); a=-a;}if(a>=10) Out(a/10);putchar(a%10+'0');
}
const int N=1005;
//Code begin...char Ma[N<<1];
int Mp[N<<1];
char a[15]={'A', 'H', 'I', 'M', 'O', 'T', 'U', 'V','W','X','Y'};void Manacher(char *s, int len){int l=0;Ma[l++]='$'; Ma[l++]='#';FOR(i,0,len-1) Ma[l++]=s[i], Ma[l++]='#';Ma[l]=0;int mx=0, id=0;FOR(i,0,l-1) {Mp[i]=mx>i?min(Mp[2*id-i],mx-i):1;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=0, ans=0;FOR(i,0,len-1) {bool mark=false;FOR(j,0,10) if (s[i]==a[j]) {mark=true; break;}if (!mark) {if (now<=i-1) {Manacher(s+now,i-now);FOR(k,0,2*(i-now)+1) ans=max(ans,Mp[k]-1);}now=i+1;}}if (now<=len-1) {Manacher(s+now,len-now);FOR(k,0,2*(len-now)+1) ans=max(ans,Mp[k]-1);}printf("%d\n",ans);}return 0;
}

View Code

J.Lazy Physics Cat

数学水题。

K.Owl Geeks

数学水题。

M.Make Cents?

水题。

转载于:https://www.cnblogs.com/lishiyao/p/7219652.html

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