Week SP1: 2014/11/2

I wrote it until now, but I still haven't seen it in the middle ..

A: Aizu 0009 prime number: Prime Number filtering. Note that the memory may burst !!.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int N=1e7;int vis[N];void initPrime(){    int num = 0, m = sqrt (N + 0.5);    REPF(i,1,N)   vis[i]=1;    for (int i = 2; i <= m; ++i)        if (vis[i] == 1)            for (int j = i * i; j <= N; j += i) vis[j] = 0;    vis[1]=0;    for(int i=2;i<=N;i++)         vis[i]+=vis[i-1];//    for(int i=2;i<=10;i++)//        cout<<"233  "<<sum[i]<<endl;}int main(){    int n;    initPrime();    while(~scanf("%d",&n))        printf("%d\n",vis[n]);    return 0;}

B: Aizu 2224 save your cat...

C: Cf 250a paper work

The number of negative numbers in each consecutive segment cannot exceed 2. Messing around.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )int a[110],n;int num[110];int main(){    while(~scanf("%d",&n))    {        REPF(i,1,n)          scanf("%d",&a[i]);        CLEAR(num,0);        int cnt=0;        int l=0;        int m=0;        REPF(i,1,n)        {            m++;            if(a[i]<0)               cnt++;            if(cnt==2)            {                cnt=0;int j;                for(j=i+1;j<=n;j++)                {                    if(a[j]<0)  break;                    else  m++;                }                num[l++]=m;                i=j-1;                m=0;            }            if(i==n&&cnt==1)  num[l++]=m;        }        if(l==0)        {            printf("%d\n%d\n",1,n);            continue;        }        printf("%d\n",l);        REP(i,l)           printf(i==l-1?"%d\n":"%d ",num[i]);    }}

D: Cf 126B password:

The longest string is its prefix and suffix, which is the same as a string in the middle.

KMP practices:

# Include <iostream> # include <cstdio> # include <cstring> # define LMT 1000005 using namespace STD; int hash [LMT], next [LMT], Len; char STR [LMT]; void Init (void) {int I = 0, j =-1; next [0] =-1; while (I <Len) {If (j =-1 | STR [I] = STR [J]) {I ++; j ++; next [I] = J ;} else J = next [J];} for (I = 0; I <Len; I ++) {// cout <"2333" <next [I] <Endl; hash [next [I] ++ ;}} int main () {int I; scanf ("% s", STR); Len = strlen (STR); Init (); I = Len; while (next [I]> 0) // corresponds to AAA. In this case, {If (hash [next [I]) {for (Int J = 0; j <next [I]; j ++) printf ("% C", STR [J]); printf ("\ n"); Return 0;} I = next [I];} printf ("just a legend \ n"); Return 0 ;}

F: Cf 303 D biridian Forest

Question: When the maze reaches the exit, there is a duel in the middle. Ask for the minimum number of fights. Reverse BFs.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int maxn=1100;char mp[maxn][maxn];int val[maxn][maxn];int dis[maxn][maxn];int vis[maxn][maxn];int dr[4][2]={{-1,0},{1,0},{0,-1},{0,1}};int n,m,ex,ey;void BFS(){    pair<int,int>st;    CLEAR(vis,0);    vis[ex][ey]=1;    dis[ex][ey]=1;    queue<pair<int,int> >q;    int dd=0x7fffff;    int num=0;    q.push(make_pair(ex,ey));    while(!q.empty())    {        st=q.front();        q.pop();        if(dis[st.first][st.second]>dd)            break;        REP(i,4)        {            int xx=st.first+dr[i][0];            int yy=st.second+dr[i][1];            if(mp[xx][yy]!='T'&&!vis[xx][yy]&&xx>=0&&xx<n&&yy>=0&&yy<m)            {                dis[xx][yy]=dis[st.first][st.second]+1;                vis[xx][yy]=1;                if(dis[xx][yy]<=dd)                    num+=val[xx][yy];                if(mp[xx][yy]=='S')                    dd=dis[xx][yy];                q.push(make_pair(xx,yy));            }        }    }    printf("%d\n",num);}int main(){    while(~scanf("%d%d",&n,&m))    {        REP(i,n)           scanf("%s",mp[i]);        CLEAR(val,0);        REP(i,n)        {            REP(j,m)            {                if(mp[i][j]=='E')                {                    ex=i;                    ey=j;                }                if(mp[i][j]>='0'&&mp[i][j]<='9')                    val[i][j]=mp[i][j]-'0';            }        }        BFS();    }    return 0;}

G CF 215 D hot days.

Long long is infinite in the middle, and the extreme values are obtained at both ends.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int maxn=1e5+100;LL t[maxn],T[maxn],x[maxn],cost[maxn];int n,m;LL solve(int f,LL xx){    int l=1,r=m/xx+(m%xx?1:0);    LL maxnn;    if(m<=xx)   maxnn=cost[f];    else   maxnn=cost[f]+x[f]*m;    if((LL)r*xx<m)        maxnn=min(maxnn,(LL)r*cost[f]+((LL)(m-xx*r)+xx)*x[f]);    else        maxnn=min(maxnn,(LL)r*cost[f]);    return maxnn;}int main(){    while(~scanf("%d%d",&n,&m))    {         LL sum=0;         REPF(i,1,n)            scanf("%I64d%I64d%I64d%I64d",&t[i],&T[i],&x[i],&cost[i]);         REPF(i,1,n)         {             if(T[i]-t[i]>0)             {                LL xx=T[i]-t[i];                sum+=solve(i,xx);             }             else                sum+=((LL)m*x[i]+cost[i]);         }         printf("%I64d\n",sum);    }    return 0;}

I: HDU 1353/poj 2581 is a violent water problem. I have been thinking about how to record the path. In fact, I started to think of violence and did not dare to write it. C ++ is required for poj.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )int main(){    int num1,num2,num3,num4;    int a,b,c,d;double x;    while(~scanf("%lf%d%d%d%d",&x,&num1,&num2,&num3,&num4))    {        int n=(int)(x*100);//        cout<<n<<endl;        int flag=0;        for(int i=0;i<=num4;i++)        {            for(int j=0;j<=num3;j++)            {                for(int k=0;k<=num2;k++)                {                    for(int l=0;l<=num1;l++)                    {                        if(i+5*j+10*k+25*l==n)                        {                            a=l;b=k;c=j;d=i;                            flag=1;                            break;                        }                    }                    if(flag)  break;                }                if(flag)   break;            }            if(flag)  break;        }        if(flag)   printf("%d %d %d %d\n",a,b,c,d);        else   printf("NO EXACT CHANGE\n");    }    return 0;}

J: HDU 1595 find the longest of the shortest

Dijkstra calculates the maximum value of the split edge record after record the precursor.

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>typedef long long LL;using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int maxn=1100;int mp[maxn][maxn];int vis[maxn],dis[maxn];int pre[maxn];int n,m;void dijkstra(int x){    int pos;    CLEAR(vis,0);    REPF(i,1,n)       dis[i]=mp[1][n];    dis[1]=0;//    vis[1]=1;    REPF(i,1,n)    {        pos=-1;        REPF(j,1,n)        {            if(!vis[j]&&(pos==-1||dis[pos]>dis[j]))                pos=j;        }        vis[pos]=1;        REPF(j,1,n)        {            if(!vis[j]&&dis[j]>dis[pos]+mp[pos][j])            {                dis[j]=dis[pos]+mp[pos][j];                if(x)     pre[j]=pos;            }        }    }}int main(){    int u,v,w;    while(~scanf("%d%d",&n,&m))    {        CLEAR(mp,0x3f3f3f);//        REPF(i,1,n)  mp[i][i]=0;        CLEAR(pre,-1);        while(m--)        {            scanf("%d%d%d",&u,&v,&w);            if(mp[u][v]>w)   mp[u][v]=mp[v][u]=w;//            cout<<mp[u][v]<<endl;        }        dijkstra(1);//        cout<<"2333  "<<dis[n]<<endl;        int dd=dis[n];        for(int i=n;i!=1;i=pre[i])        {            int t=mp[i][pre[i]];//            cout<<"666  "<<endl;            mp[i][pre[i]]=mp[pre[i]][i]=0x3f3f3f;            dijkstra(0);            if(dis[n]>dd)                dd=dis[n];            mp[i][pre[i]]=mp[pre[i]][i]=t;        }        printf("%d\n",dd);    }    return 0;}

Week SP1: 2014/11/2