《2019 Multi-University Training Contest 3》

Fansblog:

虽然想到了威尔逊定理,但是觉得用不上。

这里有个结论,那就是相邻的质数之间的最大距离 <= 1e5 + 5。

有了这个结论,我们就能暴力找到Q。

然后根据威尔逊定理 (p - 1)! mod p = p - 1。

那么(p - 1) ! = (p - 1) * (p - 2) * .... Q * ... 1= p - 1.

所以Q! = (p - 1) / ((p - 1) * (p - 2) * ... (Q + 1) )

但是这里因为p最大为1e14,所以单纯乘法可能会爆longlong。

所以对于每个乘法都要用快速乘。

《2019 Multi-University Training Contest 3》
// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 3e5 + 5;
const int M = 1e6 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;

bool check(LL x) {
    LL m = sqrt(x);
    for(int i = 2;i <= m;++i) {
        if(x % i == 0) return false;
    }
    return true;
}
LL quick_c(LL a,LL b,LL mod)
{
    LL re = 0;
    while(b)
    {
        if(b & 1) re = (re + a) % mod;
        b >>= 1;
        a = (a << 1) % mod;
    }
    return re;
}
LL quick_mi(LL a,LL b,LL mod) {
    LL re = 1;
    while(b) {
        if(b & 1) re = quick_c(re,a,mod);
        a = quick_c(a,a,mod);
        b >>= 1;
    }
    return re;
}
int main() {
    int ca;scanf("%d",&ca);
    while(ca--) {
        LL p;scanf("%lld",&p);
        LL ans = p - 1,up;
        for(LL i = p - 1;;--i) {
            if(check(i)) {
                up = i;
                break;
            }
        }
        for(LL i = up + 1;i <= p - 1;++i) ans = quick_c(ans,quick_mi(i,p - 2,p),p);
        printf("%lld\n",ans);
    }

   // system("pause");
    return 0;
}
View Code

Find the answer:

这题个人觉得比较直白,很显然最优的取法是排序之后取最大的前缀和。

那么这点我们可以建一棵类似权值树的树,上面的节点的值代表排序后的a[L]。

然后每次查询区间最大的和即可。

注意的是这里不能对值离散化,也就是相同的点要看成不同的,这样方便计算个数。

但是这题的输出逆天了,必须要每个后面都带一个空格,不然就报pe...

《2019 Multi-University Training Contest 3》
// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N = 2e5 + 5;
const int M = 4e5 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;

int a[N],b[N],ans[N];
struct QT{int id,x;}p[N];
struct Node{int L,r,num;LL sum;}node[M << 2];
bool cmp(QT a,QT b) {
    return a.x < b.x;
}
void Pushup(int idx) {
    node[idx].sum = node[idx << 1].sum + node[idx << 1 | 1].sum;
    node[idx].num = node[idx << 1].num + node[idx << 1 | 1].num;
}
void build(int L,int r,int idx) {
    node[idx].L = L,node[idx].r = r,node[idx].sum = node[idx].num = 0;
    if(L == r) {
        return ;
    }
    int mid = (L + r) >> 1;
    build(L,mid,idx << 1);
    build(mid + 1,r,idx << 1 | 1);
}
void update(int x,LL val,int idx) {
    if(node[idx].L == node[idx].r) {
        node[idx].sum = val;
        node[idx].num = 1;
        return ;
    }
    int mid = (node[idx].L + node[idx].r) >> 1;
    if(mid >= x) update(x,val,idx << 1);
    else update(x,val,idx << 1 | 1);
    Pushup(idx);
}
int query(LL up,int idx) {
    if(node[idx].sum <= up) {
        return node[idx].num;
    }
    int num = 0;
    if(node[idx << 1].sum <= up) {
        num += node[idx << 1].num;
        num += query(up - node[idx << 1].sum,idx << 1 | 1);
    }
    else num += query(up,idx << 1);
    return num;
}
int main() {
    int ca;scanf("%d",&ca);
    while(ca--) {
        int n,m;scanf("%d %d",&n,&m);
        for(int i = 1;i <= n;++i) scanf("%d",&a[i]),p[i].id = i,p[i].x = a[i];
        sort(p + 1,p + n + 1,cmp);
        for(int i = 1;i <= n;++i) b[p[i].id] = i;
        build(1,n,1);
        for(int i = 1;i <= n;++i) {
            int ma = query(m - a[i],1);
            ans[i] = i - 1 - ma;
            update(b[i],a[i],1);
        }   
        for(int i = 1;i <= n;++i) printf("%d ",ans[i]);
        printf("\n");
    }

    //system("pause");
    return 0;
}
View Code

K Subsequence:

这题一直想的是怎么dp,居然是用费用流做的。

但是建模之后确实就很好理解了。

这里的一个问题就是普通费用流会被卡时间。

先把spfa换成dij,复杂度依旧不够,要再上原始对偶优化才行。

《2019 Multi-University Training Contest 3》
/ Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N = 4e3 + 5;
const int M = 4e6 + 5;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e9
#define dbg(ax) cout << "now this num is " << ax << endl;

int a[N];
struct edge {
    int to, capacity, cost, rev;
    edge() {}
    edge(int to, int _capacity, int _cost, int _rev) :to(to), capacity(_capacity), cost(_cost), rev(_rev) {}
};
struct Min_Cost_Max_Flow {
    int V, H[N + 5], dis[N + 5], PreV[N + 5], PreE[N + 5];
    vector<edge> G[N + 5];
    //调用前初始化
    void Init(int n) {
        V = n;
        for (int i = 0; i <= V; ++i)G[i].clear();
    }
    //加边
    void add(int from, int to, int cap, int cost) {
        G[from].push_back(edge(to, cap, cost, G[to].size()));
        G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
    }
    //flow是自己传进去的变量,就是最后的最大流,返回的是最小费用
    pii Min_cost_max_flow(int s, int t, int f) {
        int flow = 0;
        int res = 0; fill(H, H + 1 + V, 0);
        while (f) {//f - 可以达到的最大的最大流
            priority_queue <pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
            fill(dis, dis + 1 + V, INF);
            dis[s] = 0; q.push(pair<int, int>(0, s));
            while (!q.empty()) {
                pair<int, int> now = q.top(); q.pop();
                int v = now.second;
                if (dis[v] < now.first)continue;
                for (int i = 0; i < G[v].size(); ++i) {
                    edge& e = G[v][i];
                    if (e.capacity > 0 && dis[e.to] > dis[v] + e.cost + H[v] - H[e.to]) {
                        dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
                        PreV[e.to] = v;
                        PreE[e.to] = i;
                        q.push(pair<int, int>(dis[e.to], e.to));
                    }
                }
            }
            if (dis[t] == INF)break;
            for (int i = 0; i <= V; ++i)H[i] += dis[i];
            int d = f;
            for (int v = t; v != s; v = PreV[v])d = min(d, G[PreV[v]][PreE[v]].capacity);
            f -= d; flow += d; res += d*H[t];
            for (int v = t; v != s; v = PreV[v]) {
                edge& e = G[PreV[v]][PreE[v]];
                e.capacity -= d;
                G[v][e.rev].capacity += d;
            }
        }
        return pii{flow,res};
    }
};
Min_Cost_Max_Flow MCMF;
/*
i - st[i,n] ed [i + n,2 * n]
s - 2 * n + 3,t - st - 2 * n + 1,ed - 2 * n + 2
*/
int main() {
    int ca;scanf("%d",&ca);
    while(ca--) {
        int n,k;scanf("%d %d",&n,&k);
        MCMF.Init(2 * n + 2);
        for(int i = 1;i <= n;++i) scanf("%d",&a[i]);
        int s = 0,t = 2 * n + 2;
        for(int i = 1;i <= n;++i) {
            MCMF.add(s,i,1,0);   
            MCMF.add(i,i + n,1,-a[i]);
            MCMF.add(i + n,2 * n + 1,1,0);
        }
        MCMF.add(2 * n + 1,2 * n + 2,k,0);
        for(int i = 1;i <= n;++i) {
            for(int j = i + 1;j <= n;++j) {
                if(a[j] >= a[i]) MCMF.add(i + n,j,1,0);
            }
        }
        pii ans = MCMF.Min_cost_max_flow(s,t,INF);
        printf("%d\n",-ans.second);
    
    }

   // system("pause");
    return 0;
}
View Code

 

《2019 Multi-University Training Contest 3》

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