2019 Multi-University Training Contest 3

Solved	Pro.ID	Title	Ratio(Accepted / Submitted)	知识点
	1001	Azshara's deep sea	6.67%(6/90)	凸包
已补	1002	Blow up the city	31.19%(126/404)	支配树/容斥
	1003	Yukikaze and Demons	16.67%(4/24)	分治/同余
	1004	Distribution of books	19.29%(93/482)	dp/平衡树/离散化后权值线段树维护/二分
	1005	Easy Math Problem	9.30%(4/43)	杜教筛
	1006	Fansblog	21.17%(671/3170)	威尔逊定理 + 质数的密度分布
已补	1007	Find the answer	11.24%(508/4521)	线段树
	1008	Game	16.08%(23/143)	三维莫队
队友已补	1009	K Subsequence	5.81%(87/1497)	费用流
	1010	Sindar's Art Exhibition	24.66%(18/73)	树链剖分
	1011	Squrirrel	13.86%(46/332)	dfs

1002

https://www.cnblogs.com/1625--H/p/11286941.html

1006

暴力找小于P的质数Q，然后依次乘逆元

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll mul(ll a,ll b,ll p){
    ll res = 0;
    for(;b;b>>=1){
        if(b & 1)res = (res + a)%p;
        a = (a+a)%p;
    }
    return res;
}
ll pow_mod(ll a,ll b,ll p){
    ll res = 1;
    for(;b;b>>=1){
        if(b & 1)res = mul(res,a,p);
        a = mul(a,a,p);
    }
    return res;
}
bool isP(ll n){
    for(ll i=2;i*i<=n;i++){
        if(n % i == 0)return false;
    }
    return true;;
}
int main(){
    int T;scanf("%d",&T);
    while(T--){
        ll q;scanf("%lld",&q);
        ll i;
        for(i=q-1;;i--){
            if(isP(i)){
                break;
            }
        }
        //cout<<i<<endl;
        ll res = 1;
        for(ll j=i+1;j<=q-2;j++){
            res =  mul(res,pow_mod(j,q-2,q),q);
        }
        printf("%lld
",res);
    }
}

1007

单点修改线段树

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+10;
typedef long long ll;
ll sum[N];
int Q,w[N],id[N],idx[N];
struct SegTree{
    int l,r;
    ll sum;
    int cnt;
}t[4*N];
bool cmp(int x,int y){
    return w[x] < w[y];
}
void build(int p,int l,int r){
    t[p].l = l;t[p].r = r;
    if(l == r){
        t[p].sum = w[id[l]];
        t[p].cnt = 1;
        return;
    }
    int mid = l + r >> 1;
    build(p*2,l,mid);
    build(p*2+1,mid+1,r);
    t[p].sum = t[p*2].sum + t[p*2+1].sum;
    t[p].cnt = t[p*2].cnt + t[p*2+1].cnt;
}
int query(int p,int mx){
    if(t[p].sum <= mx)return t[p].r;
    if(t[p].l == t[p].r){
        if(t[p].sum <= mx)return t[p].r;
        return t[p].l;
    }
    if(t[p*2].sum > mx)return query(p*2,mx);
    return query(p*2+1,mx-t[p*2].sum);
}
int query_num(int p,int l,int r){
    if(t[p].l >= l && t[p].r <= r)return t[p].cnt;
    int mid = t[p].l + t[p].r >> 1;
    int res = 0;
    if(mid >= l)res = query_num(p*2,l,r);
    if(mid < r)res += query_num(p*2+1,l,r);
    return res;
}
void change_val(int p,int pos,int val){
    if(t[p].l == t[p].r && t[p].l == pos){
        t[p].sum = 0;return;
    }
    int mid = t[p].l + t[p].r >> 1;
    if(mid >= pos)change_val(p*2,pos,val);
    if(mid < pos)change_val(p*2+1,pos,val);
    t[p].sum = t[p*2].sum + t[p*2+1].sum;
}
void change_cnt(int p,int pos){
    if(t[p].l == pos && t[p].r == pos){
        t[p].cnt = 0;
        return;
    }
    int mid = t[p].l + t[p].r >> 1;
    if(mid >= pos)change_cnt(p*2,pos);
    if(mid < pos)change_cnt(p*2+1,pos);
    t[p].cnt = t[p*2].cnt + t[p*2+1].cnt;
}
int ans[N],n,m;
int main(){
    scanf("%d",&Q);
    while(Q--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++){
            scanf("%d",&w[i]);
            id[i] = i;
        }
        sort(id+1,id+1+n,cmp);
        for(int i=1;i<=n;i++)idx[id[i]] = i;
        for(int i=1;i<=n;i++)sum[i] = sum[i-1] + w[id[i]];
        build(1,1,n);
        for(int i=n;i>=1;i--){
            change_val(1,idx[i],w[i]);
            change_cnt(1,idx[i]);
            int mx = m - w[i];
            int pos = query(1,mx)-1;
            int cnt = query_num(1,1,pos);
            ans[i] = i - cnt - 1;
        }
        for(int i=1;i<=n;i++){
            printf("%d ",ans[i]);
        }
        puts("");
    }
    return 0;
}