Harry And Physical Teacher

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 38 Accepted Submission(s): 34

Problem Description

made a completely elastic collision with a car moving towards the same direction with a speedV(V<V BestCoder Round #14

)

,
and the car far outweighs the ball, what was the speed of the small ball after the collision?

This problem was so difficult that Harry hasn't figure out how to solve it. Therefore, he asks you for help. Could you do him this favor?

Input

They are several test cases, you should process to the end of file.

There are two integers V BestCoder Round #14

and V

for each test case. All the integers are 32-bit signed non-negative integers.

Output

For each test case, just output one line that contains an integer indicate the speed of the small ball after the collision.

Sample Input

0 10

Sample Output

-10

Source

field=problem&key=BestCoder+Round+%2314&source=1&searchmode=source">BestCoder Round #14

Recommend

heyang | We have carefully selected several similar problems for you: 5068

pid=5067">5067

pid=5066">5066

5065 5064

题意：

a,b两个球a的速度我va。

b的速度为vb。va>vb.va,vb同向且a在b的后面。

b的质量远大于a。

思路：

a肯定会撞b的。

正规的思路应该是动量守恒。机械能守恒。

然后blabla。我的思路是既然b的质量远大于a。那么b能够看做是一堵墙。而a撞墙的速度就是va-vb咯。因为弹性碰撞。撞墙后速度肯定是反向啦。所以速度就变为-(va-vb)

可是这个速度是相对墙的。所以转化为绝对速度就为-(va-vb)+vb。

具体见代码：

#include<algorithm>

#include<iostream>

#include<string.h>

#include<stdio.h>

using namespace std;

const int INF=0x3f3f3f3f;

const int maxn=100010;

typedef long long ll;

int main()

{

    int v1,v2;

    while(~scanf("%d%d",&v1,&v2))

    {

        printf("%d\n",2*v1-v2);

    }

    return 0;

}

Harry And Dig Machine

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 170 Accepted Submission(s): 50

Problem Description

As we all know, Harry Porter learns magic at Hogwarts School. However, learning magical knowledge alone is insufficient to become a great magician. Sometimes, Harry also has to gain knowledge from other certain subjects, such as
language, mathematics, English, and even algorithm.

Dumbledore, the headmaster of Hogwarts, is planning to construct a new teaching building in his school. The area he selects can be considered as an n*m grid, some (but no more than ten) cells of which might contain stones. We should remove the stones there
in order to save place for the teaching building. However, the stones might be useful, so we just move them to the top-left cell. Taking it into account that Harry learned how to operate dig machine in Lanxiang School several years ago, Dumbledore decides
to let him do this job and wants it done as quickly as possible. Harry needs one unit time to move his dig machine from one cell to the adjacent one. Yet skilled as he is, it takes no time for him to move stones into or out of the dig machine, which is big
enough to carry infinite stones. Given Harry and his dig machine at the top-left cell in the beginning, if he wants to optimize his work, what is the minimal time Harry needs to finish it?

Input

They are sever test cases, you should process to the end of file.

For each test case, there are two integers n and m.(1≤n,m≤50) BestCoder Round #14

.

The next n line, each line contains m integer. The j-th number of
i BestCoder Round #14

line a[i][j] means there are a[i][j] stones on the
j BestCoder Round #14

cell of the i BestCoder Round #14

line.( 0≤a[i][j]≤100 BestCoder Round #14

, and no more than 10 of a[i][j] will be positive integer).

Output

For each test case, just output one line that contains an integer indicate the minimal time that Harry can finish his job.

Sample Input

Sample Output

4

4

Source

field=problem&key=BestCoder+Round+%2314&source=1&searchmode=source">BestCoder Round #14

Recommend

heyang | We have carefully selected several similar problems for you:

pid=5068">5068

pid=5067">5067

5065 5064

pid=5062">5062

题意：

哈利珀特学挖掘机了。那么问题来了。。

。。--||。一个n*m(50*50)的格子。一些格子数字大于0个数不超过10(看到这就懂了).你開始在左上角。

然后你要经过全部这些数字大于0的格子然后回到左上角。

然后问你最少须要多少步完毕这个目标。每步能够往四个方向走一步。

思路：

因为数字大于0的格子不超过10所以能够壮压。

这就是经典的TSP问题啦。

先对特殊格子编号然后bfs计算两两特殊格子的距离(步数).然后就TSP了。dp[i][s]表示眼下在i走到的格子状态为s。

具体见代码:

#include<algorithm>

#include<iostream>

#include<string.h>

#include<stdio.h>

#include<queue>

using namespace std;

const int INF=0x3f3f3f3f;

const int maxn=100010;

typedef long long ll;

int maze[55][55],hs[55][55],vis[55][55];

int dp[13][1<<12],dis[15][15],base[15];

int sx[55],sy[55],cnt,n,m;

int dx[]={0,-1,0,1};

int dy[]={-1,0,1,0};

struct node

{

    int x,y,d;

};

queue<node> q;

void bfs(int id)

{

    while(!q.empty())

        q.pop();

    node tt,nt;

    memset(vis,0,sizeof vis);

    tt.x=sx[id],tt.y=sy[id];

    dis[id][id]=tt.d=0;

    vis[tt.x][tt.y]=1;

    q.push(tt);

    int nx,ny,i;

    while(!q.empty())

    {

        tt=q.front();

        q.pop();

        for(i=0;i<4;i++)

        {

            nx=tt.x+dx[i];

            ny=tt.y+dy[i];

            if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!vis[nx][ny])

            {

                vis[nx][ny]=1;

                nt.x=nx,nt.y=ny,nt.d=tt.d+1;

                if(hs[nx][ny]!=-1)

                    dis[id][hs[nx][ny]]=nt.d;

                q.push(nt);

            }

        }

    }

}

int main()

{

    int i,j,k,lim;

    base[0]=1;

    for(i=1;i<15;i++)

        base[i]=base[i-1]<<1;

    while(~scanf("%d%d",&n,&m))

    {

        cnt=0;

        memset(hs,-1,sizeof hs);

        for(i=1;i<=n;i++)

        {

            for(j=1;j<=m;j++)

            {

                scanf("%d",&maze[i][j]);

                if(maze[i][j]||(i==1&&j==1))

                {

                    sx[cnt]=i,sy[cnt]=j;

                    hs[i][j]=cnt++;

                }

            }

        }

        for(i=0;i<cnt;i++)

            bfs(i);

        memset(dp,0x3f,sizeof dp);

        dp[0][0]=0;

        lim=1<<cnt;

        for(i=0;i<lim;i++)

        {

            for(j=0;j<cnt;j++)

            {

                if(dp[j][i]==INF)

                    continue;

                for(k=0;k<cnt;k++)

                {

                    if(i&base[k])

                        continue;

                    dp[k][i|base[k]]=min(dp[k][i|base[k]],dp[j][i]+dis[j][k]);

                }

            }

        }

        printf("%d\n",dp[0][lim-1]);

    }

    return 0;

}

Harry And Math Teacher

Time Limit: 5000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 170 Accepted Submission(s): 48

Problem Description

As we all know, Harry Porter learns magic at Hogwarts School. However, learning magical knowledge alone is insufficient to become a great magician. Sometimes, Harry also has to gain knowledge from other certain subjects, such as language,
mathematics, English, and even algorithm.

In Hogwarts, there is a tall castle in which all teachers live. The castle is rather special. In every floor there are two doors, behind each of them there existing two stairs to the next floor’s two doors. And if you are at the i-th floor’s j-th door , then
you can just go to the next floor from this door. However, something even more interesting (or we can say "magic") can happen to the stairs: sometimes they break into pieces (making it impossible to go to the next floor), and sometimes the fragments can joint
together and become the whole stair again. Now suppose Harry is in the a-th floor (you know, Harry is the hero, so he lives in the teachers’ building somehow), and his math teacher b-th floor. Sometimes the math teacher will call Harry to his room. Facing
these magic stairs, Harry gets puzzled how he can go to see the math teacher. Can you help Harry figure out how many ways exactly he can choose without going backwards? You can assume that the change of the stairs will not happen when Harry is on his way.
Harry can begin at any doors in floor a and he can end at any doors in floor b. And as Harry want to arrive as soon as possible, so he can not go back to the past. And at the beginning all the stairs are intact. And the answer may be too large, you should
output the answer mod 1000000007.

Input

They are sever test cases, you should process to the end of file.

For each test case, there are two integers n and m(2≤n≤50000,1≤m≤50000) BestCoder Round #14

in the first line, indicate the number of the castle’s layers and the number of queries. And the following m lines, each line contains three or four integers. If the first integer op equals 0, there are two integers a and b (1≤a<b≤n BestCoder Round #14

)
follow, indicate the position of Harry and math teacher. Otherwise, there are three integers
x,y,z(1≤x<n,1≤y,z≤2) BestCoder Round #14

follow,
it means that the stair between the x BestCoder Round #14

floor’s y

door and the (x+1) BestCoder Round #14

floor’s z-th door changes its state(if it is intact, then it breaks else it joints).

Output

For each query, if op equals 0, you should output one line that contains an integer indicates the number of ways from
a BestCoder Round #14

floor to the b BestCoder Round #14

floor.

Sample Input

Sample Output

8

6

Source

field=problem&key=BestCoder+Round+%2314&source=1&searchmode=source">BestCoder Round #14

Recommend

heyang | We have carefully selected several similar problems for you:

pid=5068" target="_blank">5068

5065 5064 5062

pid=5061" target="_blank">5061

题意：

有一栋楼。

每层楼有两扇门。

每扇门后面都有两条通向上一层楼的楼梯。楼的层数为n(5e4)。然后m个操作。

0 a b 询问a楼到b楼有多少种走法。

a<b。

1 x y z 改变x层楼y们到x+1层楼z门的楼梯状态(能够走。不能够走)。

思路：

c题赛后。读了下题。发现是大水题。

我们能够用一个2*2的矩阵来表示x层楼到x+1层楼的楼梯状态。1表示能够走。

0表示不能走。然后线段树维护区间乘积即可了。

仅仅是如今的线段树的结点为一个2*2的矩阵罢了。

单点更新区间查询。

具体见代码：

#include<algorithm>

#include<iostream>

#include<string.h>

#include<stdio.h>

using namespace std;

const int INF=0x3f3f3f3f;

const int maxn=50050;

const int mod=1e9+7;

typedef long long ll;

#define lson L,mid,ls

#define rson mid+1,R,rs

struct Matrix

{

    int val[3][3];

    Matrix(){ memset(val,0,sizeof val);}

    inline Matrix operator *(const Matrix &op)

    {

        Matrix tt;

        int i,j,k;

        for(i=1;i<=2;i++)

            for(j=1;j<=2;j++)

                for(k=1;k<=2;k++)

                    tt.val[i][j]=(tt.val[i][j]+(ll)val[i][k]*op.val[k][j]%mod)%mod;

        return tt;

    }

} yb[maxn<<2];

void build(int L,int R,int rt)

{

    if(L==R)

    {

        yb[rt].val[1][1]=1;

        yb[rt].val[1][2]=1;

        yb[rt].val[2][1]=1;

        yb[rt].val[2][2]=1;

        return;

    }

    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1;

    build(lson);

    build(rson);

    yb[rt]=yb[ls]*yb[rs];

}

void update(int L,int R,int rt,int p,int u,int v)

{

    if(L==R)

    {

        yb[rt].val[u][v]^=1;

        return;

    }

    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1;

    if(p<=mid)

        update(lson,p,u,v);

    else

        update(rson,p,u,v);

    yb[rt]=yb[ls]*yb[rs];

}

Matrix qu(int L,int R,int rt,int l,int r)

{

    if(l<=L&&R<=r)

        return yb[rt];

    int ls=rt<<1,rs=ls|1,mid=(L+R)>>1;

    Matrix tt;

    tt.val[1][1]=tt.val[2][2]=1;

    if(l<=mid)

        tt=tt*qu(lson,l,r);

    if(r>mid)

        tt=tt*qu(rson,l,r);

    return tt;

}

int main()

{

    int n,m,x,y,z,i,op,ans;

    while(~scanf("%d%d",&n,&m))

    {

        n--;

        build(1,n,1);

        for(i=1;i<=m;i++)

        {

            scanf("%d",&op);

            if(!op)

            {

                scanf("%d%d",&x,&y);

                Matrix tt=qu(1,n,1,x,y-1);

                ans=0;

                for(int j=1;j<=2;j++)

                    for(int k=1;k<=2;k++)

                        ans=(ans+tt.val[j][k])%mod;

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

            }

            else

            {

                scanf("%d%d%d",&x,&y,&z);

                update(1,n,1,x,y,z);

            }

        }

    }

    return 0;

}

Harry And Biological Teacher

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 26 Accepted Submission(s): 2

Problem Description

Input

They are sever test cases, you should process to the end of file.

For each test case, there are two integers n and m (1≤n≤100000,1≤m≤100000 BestCoder Round #14

)on
the first line, indicate the number of genes and the number of queries.

The next following n line, each line contains a non-empty string composed by characters 'A' , 'T' , 'C' , 'G'. The sum of all the characters is less than 100000.

Then the next following m line, each line contains two integers a and b(1≤a,b≤n BestCoder Round #14

),
indicate the query.

Output

For each query, you should output one line that contains an integer indicate the length of longest part that the gene a’s suffix and gene b’s prefix can overlap together.

Sample Input

2 1

ACCGT

TTT

1 2

4 4

AAATTT

TTTCCC

CCCGGG

GGGAAA

1 2

2 3

3 4

4 1

Sample Output

Source

BestCoder Round #14

Recommend

heyang | We have carefully selected several similar problems for you:

pid=5068">5068

5067 5065 5064

pid=5062">5062

题意:

给你n个由A,G,C,T构成的字符串。n个字符串的总长度不超过1e5.然后m个询问。

n。m<=1e5.每次询问a,b两个字符串。找到一个最长的串使得它是a的后缀。然后也是b的前缀。输出它的长度即可了。

思路：

第一想法就是后缀自己主动机。然后就开搞了。离线处理全部询问。把全部第一个串是a的b建个邻接表。然后依次处理每一个a。先对a建后缀自己主动机。然后对于串b在自己主动机上匹配就是了。匹配的最大长度的可接受态就是答案。然后一A。甚欢。比赛结束后才发现忘了记忆化。就是对每一个a链里的每一个b仅仅须要跑一遍即可了。哎。可惜发现的太晚了。还是无情的T了。

就上后再就200ms过了。。

。真是冤死了。时间复杂度。因为顶多对n各字符串都建一次后缀自己主动机。然后每次询问最多也仅仅能匹配a串那么长。

所以复杂度为2*总串长。

就O(n)咯。

具体见代码:

#include<algorithm>

#include<iostream>

#include<string.h>

#include<stdio.h>

#include<map>

using namespace std;

const int INF=0x3f3f3f3f;

const int maxn=100010;

typedef long long ll;

char txt[maxn<<1];

int st[maxn],ans[maxn];

map<int,int> mp;

struct node

{

    node *f,*ch[30];

    int len,ac;

    void init()

    {

        ac=0;

        memset(ch,0,sizeof ch);

    }

}*root,*tail,sam[maxn<<1];

int tot,len,cnt;

struct qnode

{

    qnode *next;

    int v,id;

} ed[maxn<<1],*head[maxn];

void adde(int u,int v,int id)

{

    ed[cnt].v=v;

    ed[cnt].id=id;

    ed[cnt].next=head[u];

    head[u]=&ed[cnt++];

}

void init()

{

    tot=0;

    root=tail=&sam[tot++];

    root->init();

    root->len=0;

}

void Insert(int c)

{

    node *p=tail,*np=&sam[tot++];

    np->init(),np->len=p->len+1;

    while(p&&!p->ch[c])

        p->ch[c]=np,p=p->f;

    tail=np;

    if(!p)

        np->f=root;

    else

    {

        if(p->ch[c]->len==p->len+1)

            np->f=p->ch[c];

        else

        {

            node *q=p->ch[c],*nq=&sam[tot++];

            *nq=*q;

            nq->len=p->len+1;

            q->f=np->f=nq;

            while(p&&p->ch[c]==q)

                p->ch[c]=nq,p=p->f;

        }

    }

}

int main()

{

    int i,j,n,m,len,tl,u,v;

    while(~scanf("%d%d",&n,&m))

    {

        len=cnt=0;

        for(i=1;i<=n;i++)

        {

            st[i]=len;

            scanf("%s",txt+len);

            tl=strlen(txt+len);

            len=len+tl+1;

        }

        memset(head,0,sizeof head);

        for(i=1;i<=m;i++)

        {

            scanf("%d%d",&u,&v);

            adde(u,v,i);

        }

        for(i=1;i<=n;i++)

        {

            init();

            for(j=st[i];txt[j];j++)

                Insert(txt[j]-'A');

            for(node *p=tail;p!=NULL;p=p->f)

                p->ac=1;

            mp.clear();

            for(qnode *q=head[i];q!=NULL;q=q->next)

            {

                if(mp.count(q->v))

                {

                    ans[q->id]=mp[q->v];

                    continue;

                }

                node *p=root;

                int ct=0,aa=0;

                for(j=st[q->v];txt[j];j++)

                    if(p->ch[txt[j]-'A']!=NULL)

                    {

                        ct++,p=p->ch[txt[j]-'A'];

                        if(p->ac)

                            aa=max(aa,ct);

                    }

                    else

                        break;

                ans[q->id]=aa;

                mp[q->v]=aa;

            }

        }

        for(i=1;i<=m;i++)

            printf("%d\n",ans[i]);

    }

    return 0;

}

秒客网

BestCoder Round #14

Harry And Physical Teacher

Harry And Dig Machine

Harry And Math Teacher

Harry And Biological Teacher

相关文章