Codeforces Round #426 (Div. 1) (ABCDE)

2023-02-25 22:32:24

1. 833A The Meaningless Game

大意: 初始分数为$1$, 每轮选一个$k$, 赢的人乘$k^2$, 输的人乘$k$, 给定最终分数, 求判断是否成立.

判断一下$a\cdot b$是否是立方数, 以及能否被那个立方的因子整除即可. cbrt竟然有误差, 特判了一下, 好坑

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

int main() {

    int n;

    scanf("%d", &n);

    while (n--) {

        int a, b;

        scanf("%d%d", &a, &b);

        ll x = (ll)a*b, t = cbrt(x);

        while (t*t*t>x) --t;

        while (t*t*t<x) ++t;

        puts(t*t*t==x&&a%t==&&b%t==?"Yes":"No");

    }

}

2. 833B The Bakery

大意: 给定$n$元素序列, 求划分为最多$x$段, 每段的贡献为不同元素数, 求最大贡献.

设$dp_{i,j}$为前$i$个数分成$j$段的最大值, 开$k$棵线段树, 位置$x$维护dp[x-1][j]+[x...i]的贡献即可

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

const int N = ;

int n,k,a[N],pre[N],vis[N];

int dp[N][];

//dp[i][j] = 前i个数,分成j段的最大值

//要支持区间加, 查询区间最值

struct {

    int ma[N<<],tag[N<<];

    void pd(int o) {

        if (tag[o]) {

            ma[lc]+=tag[o],tag[lc]+=tag[o];

            ma[rc]+=tag[o],tag[rc]+=tag[o];

            tag[o]=;

        }

    }

    void add(int o, int l, int r, int ql, int qr, int v) {

        if (ql<=l&&r<=qr) return ma[o]+=v,tag[o]+=v,void();

        pd(o);

        if (mid>=ql) add(ls,ql,qr,v);

        if (mid<qr) add(rs,ql,qr,v);

        ma[o]=max(ma[lc],ma[rc]);

    }

} tr[];

int main() {

    scanf("%d%d", &n, &k);

    REP(i,,n) {

        scanf("%d",a+i);

        pre[i] = vis[a[i]];

        vis[a[i]] = i;

    }

    REP(i,,n) {

        REP(j,,k-) tr[j].add(,,n,pre[i]+,i,);

        REP(j,,k) dp[i][j] = tr[j-].ma[];

        if (i==n) break;

        REP(j,,k-) tr[j].add(,,n,i+,i+,dp[i][j]);

    }

    printf("%d\n", dp[n][k]);

}

3. 833C Ever-Hungry Krakozyabra

大意: 给定$L,R$, 将区间$[L,R]$的每个数去除数位$0$后对数位进行排序, 求一共能得到多少种数.

先特判掉1e18, 所有可能得到的数最多为$\binom{18+9}{9}=1562275$, 可以直接暴力枚举每个数$x$, 判断$x$是否能在区间$[L,R]$得到即可. 判断合法性本来写的$O(18^2)$的暴力贪心, 结果TLE on test 70. 按照官方题解改成复杂度$O(10*18)$的$dfs$就过了...... 看评论区说似乎有不用暴力枚举的做法, 没太懂

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

int ans,len,flag,cnt,a[],x[],y[];

ll L, R, fac[];

int solve(int d, int l1, int l2) {

    if (!l1&&!l2||d<) return ;

    int U=l1?x[d]:,D=l2?y[d]:;

    REP(i,U,D) if (a[i]) {

        --a[i];

        if (solve(d-,l1&&x[d]==i,l2&&y[d]==i)) return ++a[i],;

        ++a[i];

    }

    return ;

}

int chk() {

    return solve(len-,,);

}

//生成所有合法的数

void dfs(int d, int pre) {

    if (d>len) return;

    REP(i,pre,) {

        ++a[i],++cnt;

        a[] = len-cnt;

        ans += chk();

        dfs(d+,i);

        --a[i],--cnt;

    }

}

int main() {

    fac[] = ;

    REP(i,,) fac[i] = fac[i-]*;

    scanf("%lld%lld",&L,&R);

    if (R==fac[]) --R, flag = ;

    if (L==fac[]) --L;

    ll t = L;

    while (t) x[len++]=t%,t/=;

    t = R, len = ;

    while (t) y[len++]=t%,t/=;

    dfs(,);

    printf("%d\n",ans);

}

5. 833E Caramel Clouds

大意: 给定$n$朵云, 第$i$朵出现时间范围$[l_i,r_i]$, 删除需要花费$c_i$, 初始时刻为$0$, 预算为$C$. 给定$m$个询问$k_i$, 求一棵需要照$k_i$时间阳光的花最快什么时间长成, 最多删除两朵云, 询问独立.

参考的这个https://www.cnblogs.com/ECJTUACM-873284962/p/7265224.html

主要思路就是依次处理每个时间段, 预处理每个时刻的最大光照时间.

#include <iostream>

#include <sstream>

#include <algorithm>

#include <cstdio>

#include <cmath>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <bitset>

#include <functional>

#include <random>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}

ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}

//head

#ifdef ONLINE_JUDGE

const int N = 1e6+;

#else

const int N = 1e2+;

#endif

int n, C, cost[N], tag[N], b[N];

pii tr1[N],tr2[N];

//树状数组维护区间最大次大

void add(int id, int x, int v) {

    x = lower_bound(b+,b++*b,x)-b;

    for (; x<=*b; x+=x&-x) {

        if (tr1[x].y==id) tr1[x].x = max(tr1[x].x,v);

        else if (tr2[x].x<v) tr2[x] = pii(v,id);

        if (tr1[x]<tr2[x]) swap(tr1[x],tr2[x]);

    }

}

int qry(int id, int x) {

    x = upper_bound(b+,b++*b,x)-b-;

    int ret = ;

    for (; x; x^=x&-x) {

        if (id==tr1[x].y) ret=max(ret,tr2[x].x);

        else ret=max(ret,tr1[x].x);

    }

    return ret;

}

int main() {

    scanf("%d%d", &n, &C);

    vector<pii> events;

    REP(i,,n) {

        int l,r;

        scanf("%d%d%d",&l,&r,cost+i);

        events.pb(pii(l,i));

        events.pb(pii(r,i));

        b[i] = cost[i];

    }

    events.pb(pii(,));

    events.pb(pii(2e9,));

    sort(events.begin(),events.end());

    sort(b+,b++n),*b=unique(b+,b++n)-b-;

    REP(i,,*b+) tr1[i].y=tr2[i].y=-;

    vector<pii> ans;

    map<pii,int> len;

    set<int> s;

    int mx = ;

    if (events[].y) s.insert(events[].y);

    for (int i=; i<events.size(); ++i) {

        int d = events[i].x-events[i-].x;

        if (d>&&s.size()<=) {

            int p = , q = ;

            if (s.size()) p = *s.begin();

            if (s.size()==) q = *s.rbegin();

            int start = -;

            if (!p) {

                start = mx;

                len[{,}] += d;

            }

            else if (!q) {

                if (cost[p]<=C) {

                    start = len[{p,}]+len[{,}];

                    //找一个q, 使得len[{p,q}]+len[{q,0}]的最大

                    int ma = tag[p]; //相交的情况

                    //不相交的情况

                    ma = max(ma, qry(p,C-cost[p]));

                    start += ma;

                    len[{p,}] += d;

                    add(p,cost[p],len[{p,}]);

                }

            }

            else if (cost[p]+cost[q]<=C) {

                start = len[{p,q}]+len[{p,}]+len[{q,}]+len[{,}];

                len[{p,q}] += d;

                tag[p] = max(tag[p], len[{p,q}]+len[{q,}]);

                tag[q] = max(tag[q], len[{p,q}]+len[{p,}]);

            }

            if (~start&&start+d>mx) {

                mx = start+d;

                ans.pb(pii(mx, events[i].x));

            }

        }

        int id = events[i].y;

        if (!id) continue;

        if (s.count(id)) s.erase(id);

        else s.insert(id);

    }

    int m;

    scanf("%d", &m);

    while (m--) {

        int k;

        scanf("%d", &k);

        auto p = lower_bound(ans.begin(),ans.end(),pii(k,));

        printf("%d\n", p->y-(p->x-k));

    }

}

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