• 2018 “百度之星”程序设计大赛


    1001 没有兄弟的舞会

    #include <bits/stdc++.h>
    
    using namespace std;
    typedef long long ll;
    const int maxn = 100005;
    int t,n;
    int a[maxn],fa[maxn];
    struct node
    {
        int to,w;
        node(){}
        node(int _to,int _w)
        {
            to=_to;w=_w;
        }
        bool operator < (const node & _node)
        {
            return w <_node.w;
        }
    };
    vector<node> M[maxn];
    
    int main()
    {
        int t;scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            for(int i=1;i<=n;i++)M[i].clear();
            for(int i=2;i<=n;i++)
            {
                scanf("%d",&fa[i]);
            }
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
            }
            for(int i=2;i<=n;i++)
            {
                M[fa[i]].push_back(node(i,a[i]));
            }
            for(int i=1;i<=n;i++)
            {
                sort(M[i].begin(),M[i].end());
            }
            long long ans1=0,ans2=0;
            long long sum=0,sum1=0;
            for(int i=1;i<=n;i++)
            {
                if(M[i].size()==0)continue;
                else if(M[i].size()==1)
                {
                    ans1 += max((ll)0,(ll)M[i][0].w);
                    ans2 += min((ll)0,(ll)M[i][0].w);
                }
                else{
                    int len = M[i].size();
                    ans1 += max((ll)0,(ll)M[i][len-1].w);
                    ans2 += min((ll)0,(ll)M[i][0].w);
                    sum = max(sum,(ll)M[i][len-2].w);
                    sum1 = min(sum1,(ll)M[i][1].w);
                }
               // cout<<ans1<<" "<<ans2<<endl;
            }
            if(a[1]<0)ans2+=a[1];
            else ans1+=a[1];
            cout<<ans1+sum<<" "<<ans2+sum1<<endl;
        }
    
        return 0;
    }
    View Code

    1002 序列期望

    #include <bits/stdc++.h>
    using namespace std;
    int l[105],r[105];
    int mx,mi;
    long long mod = 1e9 + 7;
    long long ans = 0;
    long long cnt;
    
    long long power(long long a,long long k){
        long long ret = 1;
        a = a % mod;
        while(k){
            if(k & 1) ret = ret * a % mod;
            a = a * a % mod;
            k >>= 1;
        }
        return ret;
    }
    
    long long calc(long long beg,long long end){
        if(beg > end) return 0;
        long long ret = (beg+end)*(end-beg+1)/2;
        return ret;
    }
    
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--){
            int n;
            scanf("%d",&n);
            mx = 0,mi = 0;
            ans = 0;
            long long tot = 1;
            for(int i = 1;i <= n;++i){
                scanf("%d%d",&l[i],&r[i]);
                mx = max(mx,r[i]);
                mi = max(l[i],mi);
                tot = tot * (r[i]-l[i]+1) % mod;
            }
            
            long long up,tmp,tmp2,up2;
            for(int i = mi;i <= mx;++i){
                tmp = 1;
                tmp2 = 1;
                for(int j = 1;j <= n;++j){
                    up = min(r[j],i);
                    up2 = min(i-1,r[j]);
                    tmp2 = tmp2 * calc(i-up2+1,i-l[j]+1) % mod;
                    tmp = tmp * calc(i-up+1,i-l[j]+1) % mod;
                }
                ans = ans + tmp - tmp2 + mod;
                ans %= mod;
            }
            ans = ans * power(tot,mod-2)%mod;
            printf("%I64d
    ",ans);
        }
        return 0;
    }
    View Code

    1003 带劲的and和

    #include <bits/stdc++.h>
    
    using namespace std;
    typedef long long ll;
    const int maxn = 100005;
    const int mod = 1e9+7;
    int n,m;
    int t;
    int v[maxn];
    ll ans = 0;
    int par[maxn];
    ll poww[35];
    int ta[maxn][35],tb[maxn][35];
    struct number
    {
        int x,id;
        bool operator < (const number & _number)
        {
            if(id == _number.id)
            {
                return x < _number.x;
            }
            return id < _number.id;
        }
    }tu[maxn];
    
    void init(int n)
    {
        for (int i = 0; i <= n; i++) {
            par[i] = i;
        }
    }
    int fin(int x) {
        if (par[x] == x) {
            return x;
        }
        else {
            return par[x] = fin(par[x]);
        }
    }
    
    void unite(int x, int y) {
        x = fin(x);
        y = fin(y);
        if (x == y) return;
        par[x] = y;
    
    }
    
    bool same(int x, int y) {
        return fin(x) == fin(y);
    }
    void bi(int pos,int num)
    {
        int i=0;
        while(num)
        {
            ta[pos][i]=(num%2);i++;
            num/=2;
        }
    }
    ll ask(int l,int r)
    {
    
    
        for(int i=l;i<=r;i++)
        {
            bi(i,tu[i].x);
        }
        for(int i=0;i<=31;i++)
        {
            tb[l][i]=ta[l][i];
        }
        for(int i=l+1;i<=r;i++)
        {
            for(int j=0;j<=31;j++)
            {
                tb[i][j] = tb[i-1][j] + ta[i][j];
            }
        }
    
        ll ret = 0;
    
        for(int i=l+1;i<=r;i++)
        {
            for(int j=0;j<=31;j++)
            {
                if(ta[i][j])
                    ret = (ret + (ll)(tu[i].x*poww[j])%mod*tb[i-1][j])%mod;
            }
        }
    
        ret%=mod;
    
        return ret;
    }
    int main()
    {
        poww[0]=1;
        for(int i=1;i<=31;i++)
        {
            poww[i]=poww[i-1]*2;
        }
        scanf("%d",&t);
        while(t--)
        {
            ans = 0;
            scanf("%d%d",&n,&m);
            init(n);
            for(int i=1;i<=n;i++)scanf("%d",&v[i]);
            for(int i=1;i<=m;i++)
            {
                int u,uu;
                scanf("%d%d",&u,&uu);
                unite(u,uu);
            }
            for(int i=1;i<=n;i++)
            {
                tu[i].id=fin(i);
                tu[i].x=v[i];
            }
            sort(tu+1,tu+n+1);
            int l=1,r=1;
            memset(ta,0,sizeof(ta));
    
            for(int i=2;i<=n;i++)
            {
                if(tu[i].id == tu[i-1].id)r++;
                else
                {
                    ans = (ans+ask(l,r))%mod;
                    l = r = i;
                }
                //cout<<ans<<endl;
            }
           // cout<<l<<" fjsdkfldk   "<<r<<endl;
            ans = (ans+ask(l,r))%mod;
            cout<<ans<<endl;
        }
    
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/solvit/p/9559663.html
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