百度之星2018 复赛 体验记

毒瘤卡常啊我勒个去

A 题被卡常我要骂人了

C 题写了一个常数大的做法，被卡到怀疑人生（后来重写才 AC 的mdzz）。

还好 B 和 D 比较善良

题解不想写了

直接放代码QAQ

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100005;
LL read(){
    LL x=0,f=1;
    char ch=getchar();
    while (!isdigit(ch)&&ch!=‘-‘)
        ch=getchar();
    if (ch==‘-‘)
        f=-1,ch=getchar();
    while (isdigit(ch))
        x=(x<<1)+(x<<3)+ch-48,ch=getchar();
    return x*f;
}
int n,A[N],fa[N];
int a1[N],b1[N],a2[N],b2[N];
int main(){
    int T=read();
    while (T--){
        n=read();
        memset(a1,0,sizeof a1);
        memset(a2,0,sizeof a2);
        memset(b1,0,sizeof b1);
        memset(b2,0,sizeof b2);
        for (int i=2;i<=n;i++)
            fa[i]=read();
        for (int i=1;i<=n;i++)
            A[i]=read();
        for (int i=2;i<=n;i++){
            int v=A[i],f=fa[i];
            if (v>=a1[f])
                a2[f]=a1[f],a1[f]=v;
            else
                a2[f]=max(a2[f],v);
            if (v<=b1[f])
                b2[f]=b1[f],b1[f]=v;
            else
                b2[f]=min(b2[f],v);
        }
        LL Max=0,Min=0;
        for (int i=1;i<=n;i++){
            Max+=a1[i];
            Min+=b1[i];
        }
        LL ans1=Max,ans2=Min;
        for (int i=1;i<=n;i++){
            ans1=max(ans1,Max+a2[i]);
            ans2=min(ans2,Min+b2[i]);
        }
        printf("%I64d %I64d\n",ans1+max(A[1],0),ans2+min(A[1],0));
    }
    return 0;
}

A

#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=105,V=10005,mod=1e9+7;
int T,n,L[N],R[N],inv[N];
int Pow(int x,int y){
    int ans=1;
    for (;y;y>>=1,x=1LL*x*x%mod)
        if (y&1)
            ans=1LL*ans*x%mod;
    return ans;
}
int calc(int x){
    return x*(x+1)/2;
}
int main(){
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        int MaxL=0,MaxR=0;
        for (int i=1;i<=n;i++){
            scanf("%d%d",&L[i],&R[i]);
            MaxL=max(MaxL,L[i]);
            MaxR=max(MaxR,R[i]);
            inv[i]=Pow(R[i]-L[i]+1,mod-2);
        }
        int ans=0;
        for (int i=MaxL;i<=MaxR;i++){
            int v0=1,v1=0;
            for (int j=1;j<=n;j++){
                if (R[j]<i){
                    v0=1LL*v0*(calc(i-L[j]+1)-calc(i-R[j]))%mod;
                    v1=1LL*v1*(calc(i-L[j]+1)-calc(i-R[j]))%mod;
                }
                else {
                    v1=(1LL*v1*calc(i-L[j]+1)%mod+1LL*v0%mod)%mod;
                    v0=1LL*v0*(calc(i-L[j]+1)-1)%mod;
                }
                v1=1LL*v1*inv[j]%mod;
                v0=1LL*v0*inv[j]%mod;
            }
            ans=(ans+v1)%mod;
//            printf("%d %d\n",v0,v1);
        }
        printf("%d\n",ans);
    }
    return 0;
}

B

#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=100005,mod=1e9+7;
int read(){
    int x=0;
    char ch=getchar();
    while (!isdigit(ch))
        ch=getchar();
    while (isdigit(ch))
        x=(x<<1)+(x<<3)+ch-48,ch=getchar();
    return x;
}
struct Gragh{
    static const int M=N*2;
    int cnt,y[M],nxt[M],fst[N];
    void add(int a,int b){
        y[++cnt]=b,nxt[cnt]=fst[a],fst[a]=cnt;
    }
}g;
int n,m,v[N],fa[N],id[N],p[N];
vector <int> vs[N];
int getf(int x){
    return fa[x]==x?x:fa[x]=getf(fa[x]);
}
bool cmp(int a,int b){
    return v[a]<v[b];
}
int main(){
    int T=read();
    while (T--){
        n=read(),m=read();
        for (int i=1;i<=n;i++)
            v[i]=read(),id[i]=i;
        sort(id+1,id+n+1,cmp);
        for (int i=1;i<=n;i++)
            p[id[i]]=i,fa[i]=i;
        g.cnt=0;
        memset(g.fst,0,sizeof(int)*(n+1));
        for (register int i=1;i<=m;i++){
            int a=read(),b=read();
            fa[getf(a)]=getf(b);
        }
        for (int i=1;i<=n;i++)
            vs[i].clear();
        for (int i=1;i<=n;i++)
            vs[getf(id[i])].push_back(v[id[i]]);
        int ans=0;
        for (int i=1;i<=n;i++){
            if (vs[i].size()<1)
                continue;
            for (int d=0;d<30;d++){
                int cnt=0;
                for (vector <int> :: iterator p=vs[i].begin();
                                p!=vs[i].end();p++){
                    if (!(((*p)>>d)&1))
                        continue;
                    ans=((1LL<<d)*cnt%mod*(*p)+ans)%mod;
                    cnt++;
                }
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

C

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1005,mod=1e9+7;
LL read(){
    LL x=0;
    char ch=getchar();
    while (!isdigit(ch))
        ch=getchar();
    while (isdigit(ch))
        x=(x<<1)+(x<<3)+ch-48,ch=getchar();
    return x;
}
struct Pos{
    int p[5];
}s[N*6];
int n,k,cnt,p[5],in[N*6];
vector <int> v[5][N],e[N*6];
bool check(Pos a,Pos b){
    for (int i=0;i<k;i++)
        if (a.p[i]>=b.p[i])
            return 0;
    return 1;
}
void dfs(int val,int d){
    if (d>=k){
        cnt++;
        for (int i=0;i<k;i++)
            s[cnt].p[i]=p[i];
        return;
    }
    for (int i=0;i<v[d][val].size();i++){
        p[d]=v[d][val][i];
        dfs(val,d+1);
    }
}
int q[N*6],head,tail;
int dp[N*6];
int main(){
    int T=read();
    while (T--){
        k=read(),n=read();
        for (int i=0;i<k;i++)
            for (int j=1;j<=n;j++)
                v[i][j].clear();
        for (int i=0;i<k;i++)
            for (int j=1;j<=n;j++)
                v[i][read()].push_back(j);
        cnt=0;
        for (int i=1;i<=n;i++)
            dfs(i,0);
        for (int i=1;i<=cnt;i++)
            e[i].clear();
        memset(in,0,sizeof in);
        for (int i=1;i<=cnt;i++)
            for (int j=1;j<=cnt;j++)
                if (check(s[i],s[j]))
                    e[i].push_back(j),in[j]++;
        head=tail=0;
        for (int i=1;i<=cnt;i++)
            if (!in[i])
                q[++tail]=i;
        for (int i=1;i<=cnt;i++)
            dp[i]=1;
        int ans=0;
        while (head<tail){
            int x=q[++head];
            ans=(ans+dp[x])%mod;
            for (int i=0;i<e[x].size();i++){
                int y=e[x][i];
                dp[y]=(dp[y]+dp[x])%mod;
                in[y]--;
                if (!in[y])
                    q[++tail]=y;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

D

原文地址：https://www.cnblogs.com/zhouzhendong/p/Baidu-Star-2018-Rematch.html