2019nc#4

题号	标题	已通过代码	题解	通过率	团队的状态
A	meeting	点击查看	树直径	604/2055
B	xor	点击查看	线段树维护线性基交	81/861	未通过
C	sequence	点击查看	单调栈，笛卡尔树	479/2755
D	triples I	点击查看	构造	464/2974
E	triples II	点击查看	进入讨论	35/84	未通过
F	merge	点击查看	splay，FHQ-TREE	4/37
G	tree	点击查看	进入讨论	2/43	未通过
H	RNGs	点击查看	进入讨论	1/66	未通过
I	string	点击查看	后缀数组，回文树	157/677
J	free	点击查看	分层图最短路	784/2790
K	number	点击查看	dp	859/3484

K

题意

问一个字符串中有多少个连续子串是300的倍数

思路

O（300n）的dp即可

#include <bits/stdc++.h>

using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

/**********showtime************/
            const int maxn = 1e5+;
            ll dp[maxn][];
            char str[maxn];
int main(){
            scanf("%s", str + );
            int n = strlen(str + );
//            dp[0][0] = 1;
            ll ans = ;
            for(int i=; i<n; i++) {
                int tp = str[i + ] - '';
                for(int j=; j<; j++) {
//                    debug(tp);
                    dp[i+][(j* + tp) % ] += dp[i][j];
                }
                dp[i+][tp] ++;
                ans += dp[i+][];
            }
            printf("%lld\n", ans);
            return ;
}

F merge

题意

给定一个1～n的排列，有两种操作，1）对区间 [le，mid]和[mid+1, ri] 进行一次归并排序，2）查询区间第i个位子上的值。

思路

利用fhq_tree。区间分裂操作。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include <bits/stdc++.h>

using namespace std;
#define pb push_back
#define fi first
#define se second
#define debug(x) cerr<<#x << " := " << x << endl;
#define bug cerr<<"-----------------------"<<endl;
#define FOR(a, b, c) for(int a = b; a <= c; ++ a)

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

template<typename T>
inline T read(T&x){
    x=;int f=;char ch=getchar();
    while (ch<''||ch>'') f|=(ch=='-'),ch=getchar();
    while (ch>=''&&ch<='') x=x*+ch-'',ch=getchar();
    return x=f?-x:x;
}

const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f;
const int mod = ;

/**********showtime************/

struct fhq_treap {
    static const int N = 1e5 + ;
    struct Node {
        int val, key, lc, rc, sz, mx;
    }tree[N];
    int rt, tot;
    inline void init() {
        rt = tot = ;
        tree[rt].sz = tree[rt].val = tree[rt].lc = tree[rt].rc = ;
        srand(time());
    }
    inline void update(int rt) {
        tree[rt].sz = tree[tree[rt].lc].sz +  + tree[tree[rt].rc].sz;
        tree[rt].mx = max(tree[tree[rt].lc].mx,max(tree[rt].val,tree[tree[rt].rc].mx));
    }

    void split_val(int rt, int &a, int &b, int val) {
        if(rt == ) {a = b = ; return ;}
        if(max(tree[rt].val , tree[tree[rt].lc].mx) <= val) {
            a = rt;
            split_val(tree[rt].rc, tree[a].rc, b, val);
        }
        else {
            b = rt;
            split_val(tree[rt].lc, a, tree[b].lc, val);
        }
        update(rt);
    }
    void split_sz(int rt, int &a, int &b, int sz) {
        if(rt == ) {a = b = ; return ;}
        if(tree[tree[rt].lc].sz +  > sz) {
            b = rt;
            split_sz(tree[rt].lc, a, tree[b].lc, sz);
        }
        else {
            a = rt;
            split_sz(tree[rt].rc, tree[a].rc, b, sz -  - tree[tree[rt].lc].sz);
        }
        update(rt);
    }

    void merge(int &rt, int a, int b) {
        if(a== || b==) {
            rt = a+b;
            return ;
        }
        if(tree[a].key < tree[b].key) {
            rt = a;
            merge(tree[rt].rc, tree[a].rc, b);
        }
        else {
            rt = b;
            merge(tree[rt].lc, a, tree[b].lc);
        }
        update(rt);
    }

    inline int new_node(int val) {
        tree[++tot].sz = ;
        tree[tot].val = val;
        tree[tot].lc = tree[tot].rc = ;
        tree[tot].key = rand();
        tree[tot].mx = val;
        return tot;
    }

    void ins(int &rt, int val) {
        int x = , y = , node = new_node(val);
        merge(rt, rt, node);
    }
    void delete_node(int &rt, int val) {
        int x = , y = , z = ;
        split_val(rt, x, y, val);
        split_val(x, x, z, val-);
        merge(z, tree[z].lc, tree[z].rc);
        merge(x, x, z);
        merge(rt, x, y);
    }
    inline int get_kth(int rt, int k) {
        while(tree[tree[rt].lc].sz+ != k) {
            if(tree[tree[rt].lc].sz >= k) rt = tree[rt].lc;
            else k -= tree[tree[rt].lc].sz+, rt = tree[rt].rc;
        }
        return tree[rt].val;
    }
    int get_rnk(int &rt, int val) {
        int x = , y = ;
        split_val(rt, x, y, val-);
        int tmp = tree[x].sz+;
        merge(rt, x, y);
        return tmp;
    }
    int get_pre(int &rt, int val) {
        int x = , y = ;
        split_val(rt, x,  y, val-);
        int tmp = get_kth(x, tree[x].sz);
        merge(rt, x, y);
        return tmp;
    }
    int get_scc(int &rt, int val) {
        int x = , y = ;
        split_val(rt, x, y, val);
        int tmp = get_kth(y, );
        merge(rt, x, y);
        return tmp;
    }
}t;

            const int maxn = 1e5+;
            int n,m;
            int a[maxn];
            void display(int x) {
                if(t.tree[x].lc) display(t.tree[x].lc);
                if(t.tree[x].rc) display(t.tree[x].rc);
            }
            void solve(int le, int mid ,int ri) {
                int x, y, z;
                t.split_sz(t.rt, x, z, ri);
                t.split_sz(x, x, y, mid);

                int tmp;
                int r = ;
                while(x && y) {
                    int p = t.get_kth(x, );
                    int q = t.get_kth(y, );

                    if(p > q) swap(p, q), swap(x, y);
                    t.split_val(x, tmp, x, q);
                    t.merge(r,r, tmp);
                }
                t.merge(r, r,  x);
                t.merge(r, r,  y);
                t.merge(r, r,  z);
                t.rt = r;
            }
int main(){
            t.init();
            scanf("%d%d", &n, &m);
            for(int i=; i<=n; i++) {
                scanf("%d", &a[i]);
                t.ins(t.rt, a[i]);
            }
            while(m--){
                int op; scanf("%d", &op);
                if(op == ) {
                    int x;  scanf("%d", &x);
                    printf("%d\n", t.get_kth(t.rt, x));
                }
                else {
                    int le, mid, ri;
                    scanf("%d%d%d", &le, &mid, &ri);
                    solve(le, mid, ri);
                }
            }
            return ;
}

/*
5 10
3 2 1 5 4
1 1 2 5

*/