# Part of the online competition in the ACM school competition and the acm school Competition

Source: Internet
Author: User

Part of the online competition in the ACM school competition and the acm school Competition

I have three questions in this online competition.

Vending machine

Because there are many test examples, the calculation times out every time, so you need to create a table, the recursive equation is dp [I] [j] = dp [I-1] [j] + dp [I-1] [J-1]; pay attention to the optimization of pruning, the actual 2000*1000 times out.

`#include<iostream>#include<cstdio>#include<cstdlib>#include<string.h>#include<math.h>#include<algorithm>#include<map>#include<vector>#include<queue>using namespace std;int dp[2005][1005];const int mod=(1e9+7);int main(){    //freopen("in3.txt","r",stdin);    //freopen("out3.txt","w",stdout);    dp[1][0]=1;    for(int i=2;i<=2000;i++)    {        dp[i][0]=1;        int len=i/2;        for(int j=1;j<=len;j++)        {            dp[i][j]=(dp[i-1][j]+dp[i-1][j-1]);            if(dp[i][j]>=mod)            dp[i][j]%=mod;        }    }    int n,m;    while(~scanf("%d%d",&n,&m))    {        printf("%d\n",dp[n+m][m]);    }    return 0;}`

KSS Gold Card dream 1

Map discretization + topological sorting of changes, the time limit is relatively tight, I did not expect that the OJ running speed is so slow, the local machine is less than 1 s, my pot.

`#include<iostream>#include<cstdio>#include<cstdlib>#include<string.h>#include<math.h>#include<algorithm>#include<vector>#include<queue>#include<map>using namespace std;const int MAXN=505;int g[MAXN][MAXN],in[MAXN];bool vis[MAXN];int main(){    //freopen("in1.txt","r",stdin);    int n,m,cnt;    while(~scanf("%d",&n))    {        map<string,int> mp;        cnt=1;        memset(g,0,sizeof(g));        memset(in,0,sizeof(in));        memset(vis,false,sizeof(vis));        while(n--)        {            char a[50],b[50];            //string a,b;            scanf("%s%s",a,b);            if(mp[a]==0)mp[a]=cnt++;            if(mp[b]==0)mp[b]=cnt++;            g[mp[b]][mp[a]]=1;            in[mp[a]]++;//cout<<n<<endl;        }        bool flag=false;        for(int i=1;i<cnt;i++)        {            for(int j=1;j<cnt;j++)            {                if(in[j]==0 && !vis[j])                {                    vis[j]=1;                    flag=true;                    for(int k=1;k<cnt;k++)                        if(g[j][k])                            in[k]--;                    break;                }            }            if(flag==false)                break;            flag=false;        }        scanf("%d",&m);        int num=0;        for(int i=0;i<m;i++)        {            string a;            cin>>a;            if(mp[a]>0 && vis[mp[a]]) num++;        }        if(1.0*num/m>0.7)        cout<<"Excelsior!"<<endl;        else        cout<<"KSS have a dream!"<<endl;    }    return 0;}`

KSS gold dream 2

Offline storage, sorting by distance, adding the Chair tree, maintaining the range and the number of overwrites. The result of the binary location query is that there are many, many, and many WA points, which are my pot.

`#include<iostream>#include<cstdio>#include<cstdlib>#include<string.h>#include<math.h>#include<algorithm>#include<vector>#include<queue>#include<map>#include<time.h>using namespace std;const int MAXN=100005;struct node{    int l,d,val,flag;}in[MAXN<<1];struct node1{    int x,a,b,c;}in2[MAXN];struct tree{    int l,r,cnt,flag[2];    long long val;}tree[MAXN*35];int root[MAXN<<1],pos;bool cmp1(node a,node b){    return a.d<b.d;}bool cmp2(node a,node b){    return a.l<b.l;}void pushup(int now){    tree[now].cnt=0;    tree[now].val=0;    if(tree[now].flag[0]!=-1)    {        tree[now].cnt+=tree[tree[now].flag[0]].cnt;        tree[now].val+=tree[tree[now].flag[0]].val;    }    if(tree[now].flag[1]!=-1)    {        tree[now].cnt+=tree[tree[now].flag[1]].cnt;        tree[now].val+=tree[tree[now].flag[1]].val;    }}int build(int l,int r){    int now=pos++;    tree[now].l=l;    tree[now].r=r;    tree[now].cnt=tree[now].val=0;    tree[now].flag[0]=tree[now].flag[1]=-1;    if(l==r)        return now;    int mid=(l+r)>>1;    tree[now].flag[0]=build(l,mid);    tree[now].flag[1]=build(mid+1,r);    return now;}int update(int l,int r,int v,int v1,int pre,int flag){    int now=pos++;    tree[now].l=l;    tree[now].r=r;    tree[now].flag[0]=tree[pre].flag[0];    tree[now].flag[1]=tree[pre].flag[1];    tree[now].cnt=tree[pre].cnt;    tree[now].val=tree[pre].val;    if(l==r)    {        if(flag)        {            tree[now].cnt--;            tree[now].val-=v1;        }        else        {            tree[now].cnt++;            tree[now].val+=v1;        }        return now;    }    int mid=(l+r)>>1;    if(v<=mid)        tree[now].flag[0]=update(l,mid,v,v1,tree[pre].flag[0],flag);    else        tree[now].flag[1]=update(mid+1,r,v,v1,tree[pre].flag[1],flag);    pushup(now);    return now;}long long query(int l,int r,int now,long long k){    if(l==r)    {            if(tree[now].cnt==0)                return 0;            return tree[now].val/tree[now].cnt*k;    }    int mid=(l+r)>>1;    long long ans=0;    if(k<=tree[tree[now].flag[0]].cnt)        ans+=query(l,mid,tree[now].flag[0],k);    else        ans+=tree[tree[now].flag[0]].val+query(mid+1,r,tree[now].flag[1],k-=tree[tree[now].flag[0]].cnt);    return ans;}int main(){    //freopen("in2.txt","r",stdin);    //freopen("out2.txt","w",stdout);    int n,m,x,p;    while(~scanf("%d%d%d%d",&n,&m,&x,&p))    {        long long prescore=1;        pos=0;        int l,r,d,num=1;        for(int i=1;i<=n;i++)        {                scanf("%d%d%d",&l,&r,&d);                in[num].l=l;                in[num].d=d;                in[num++].flag=0;                in[num].l=r+1;                in[num].d=d;                in[num++].flag=1;        }        for(int i=0;i<m;i++)            scanf("%d%d%d%d",&in2[i].x,&in2[i].a,&in2[i].b,&in2[i].c);        sort(in+1,in+num,cmp1);        int cnt=0;        in[num].d=-1;        for(int i=1;i<num;i++)            if(in[i].d!=in[i+1].d)                in[i].val=cnt++;            else                in[i].val=cnt;        sort(in+1,in+num,cmp2);        root[0]=build(0,cnt-1);        for(int i=1;i<num;i++)            root[i]=update(0,cnt-1,in[i].val,in[i].d,root[i-1],in[i].flag);        for(int i=0;i<m;i++)        {                int low=1,high=n*2;                int ans1=0;                while(low<=high)                {                    int mid=(low+high)>>1;                    if(in[mid].l<=in2[i].x)                    {                        ans1=mid;                        low=mid+1;                    }                    else                        high=mid-1;                }                long long nowk=((long long)in2[i].a*prescore+(long long)in2[i].b)%(long long)in2[i].c;                int flag=0;                if(prescore>p)                    flag=1;                if(ans1)                {                    if(nowk>=tree[root[ans1]].cnt)                    {                        prescore=(tree[root[ans1]].val);                        if(flag)                        prescore*=2;                    }                    else                    {                        prescore=query(0,cnt-1,root[ans1],nowk);                        if(flag)                        prescore*=2;                    }                }                else                    prescore=0;                printf("%I64d\n",prescore);        }    }    return 0;}`

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