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【数据结构】【CF1073D】 Berland Fair

发布时间:2021-04-01 08:05:20 所属栏目:安全 来源:网络整理
导读:副标题#e# Description 给定 (n) 个商店,他们围成一个圆圈,按照顺时针从 (1) 到 (n) 编号。你有 (T) 元钱,从 (1) 号点开始按照顺时针方向走,每到一个商店,只要钱够就必须买这个商店的物品。商店中物品是无限的,即多次到达可能多次购买。求

(O(n log n))

#include <ctime>
#include <cstdio>
#ifdef ONLINE_JUDGE
#define freopen(a,c)
#endif
#define rg register
#define ci const int
#define cl const long long

typedef long long ll;

namespace IPT {
    const int L = 1000000;
    char buf[L],ch = IPT::GetChar();
    }
    if (lst == '-') x = -x;
}

namespace OPT {
    int buf[120];
}

template <typename T>
inline void qw(T x,putchar('-');}
    rg int top=0;
    do {OPT::buf[++top] = int(x % 10 + '0');} while (x /= 10);
    while (top) putchar(OPT::buf[top--]);
    if (pt) putchar(aft);
}

const int maxn = 200010;
const int maxt = 400010;

int n;
ll t,ans;
int tree[maxn],MU[maxn];

struct Tree {
    Tree *ls,*rs;
    int l,r;
    ll v;
    
    inline void pushup() {
        this->v = this->ls ? (this->rs ? this->ls->v + this->rs->v : this->ls->v) : this->rs->v;
    }
};
Tree *pool[maxt],qwq[maxt],*rot;
int top;

inline int lowbit(ci x) {return x & -x;}

int check(int,ll);
void buildpool();
void build(Tree*,ci,ci);
void update(int,ci);
void update(Tree*,ci);
ll ask(Tree*,ci);
int ask(int);
int ask(Tree*,cl);

signed main() {
    freopen("1.in",stdin);
    qr(n); qr(t); ll s = 0; 
    for (rg int i = 1; i <= n; ++i) {qr(MU[i]); s+= MU[i]; update(i,1);}
    int cnt = n;
    buildpool();
    build(rot,n);
    while (cnt) {
        if (!t) break;
        ans += t / s * cnt; t %= s;
        int k = 0;
        do {
            int pre = k;
            k = check(k,t);
            if (k > n) {
                t -= ask(rot,pre + 1,n);
                ans += ask(n) - ask(pre);
                break;
            };
            update(rot,k); update(k,-1); --cnt;
            t -= ask(rot,pre,k - 1);
            s -= MU[k];
            ans += ask(k) - ask(pre);
        } while (cnt && t);
    }
    qw(ans,true);
}


int check(int pre,ll x) {
    x += ask(rot,pre);
    ll s = ask(rot,n);
    if (s <= x) return n + 1;
    return ask(rot,x);
}

void buildpool() {
    for (rg int i = 0; i < maxt; ++i) pool[i] = qwq + i;
    rot = pool[maxt - 1];  top = maxt - 2;
}

void build(Tree *u,ci l,ci r) {
    u->l = l; u->r = r;
    if (l == r) {u->v = MU[l]; return;}
    int mid = (l + r) >> 1;
    if (l <= mid) build(u->ls = pool[top--],l,mid);
    if (mid < r) build(u->rs = pool[top--],mid + 1,r);
    u->pushup();
}

void update(int x,ci v) {
    while (x <= n) {
        tree[x] += v;
        x += lowbit(x);
    }
}

void update(Tree* u,ci x) {
    if ((u->l > x) || (u->r < x)) return;
    if (u->l == u->r) {u->v = 0; return;}
    if (u->ls) update(u->ls,x);
    if (u->rs) update(u->rs,x);
    u->pushup();
}

int ask(int x) {
    int _ret = 0;
    while (x) {
        _ret += tree[x];
        x -= lowbit(x);
    }
    return _ret;
}

ll ask(Tree *u,ci r) {
    if ((u->l > r) || (u->r < l)) return 0;
    if ((u->l >= l) && (u->r <= r)) return u->v;
    return u->ls ? (u->rs ? ask(u->ls,r) + ask(u->rs,r) : ask(u->ls,r)) : ask(u->rs,r);
}

int ask(Tree *u,cl v) {
    if (u->l == u->r) return u->l;
    if (u->ls->v <= v) return ask(u->rs,v - u->ls->v);
    else return ask(u->ls,v);
}

Summary

该define int ll就要define啊= =要不然可能会fst的很惨= =

一个数对比自己小的数取模一次至少减少一半。

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