暑假集训csp提高模拟9

赛时rank15 T1 0,T2 100,T3 0,T4 0

T1，T3 都会做，然后都挂了。

恼了，挂200，不愧是我，唐

T1 大众点评

「JOISC 2014 Day1」拉面比较

简单的交互。

考虑选择相邻的两组，小的单独存一个，大的单独存一个，是比较200次

再将大的互相比较，小的互相比较，各200次

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
#include "ramen.h"
using namespace std;
void Ramen(int N){vector<int> mn,mx;// Answer(1000000,10000);if(N == 1)return Answer(0,0),void();int mxpos,mnpos;for(int i = 1;i < N; i += 2){int res = Compare(i-1,i);if(res == -1){mn.push_back(i-1);mx.push_back(i);}else{mx.push_back(i-1);mn.push_back(i);}}if(N&1) mn.push_back(N-1),mx.push_back(N-1);mxpos = mx[0];for(int i = 1;i < mx.size(); ++i){if(Compare(mx[i],mxpos) == 1) mxpos = mx[i];}mnpos = mn[0];for(int i = 1;i < mn.size(); ++i){if(Compare(mn[i],mnpos) == -1) mnpos = mn[i];}Answer(mnpos,mxpos);
}

T2 录取查询

[ABC285F] Substring of Sorted String

简单题。

我们发现当一个字符串可以是子串时当且仅当它是升序排列的，且除了第一个字符和最后一个字符，中间的字符在整个字符串的其他位置没有出现过。

权值树状数组记录一下出现次数，线段树记录一下区间极值、区间是否单调。注意一下树状数组和线段树的细节。

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;using db = double;
#ifdef LOCALFILE *InFile = infile("in.in"),*OutFile = outfile("out.out");// FILE *ErrFile=errfile("err.err");
#elseFILE *Infile = stdin,*OutFile = stdout;//FILE *ErrFile = stderr;
#endif
const int N = 1e5 + 10;
struct BIT{
private:#define lowbit(x) (x&(-x))int tree[N];
public:int mx;inline void update(int pos,int val){for(int i = pos;i <= mx;i += lowbit(i)) tree[i] += val;}inline int query(int pos){int res = 0;for(int i = pos; i;i -= lowbit(i)) res += tree[i];return res;}
}T[27];
char s[N];
int n,m;
class Segment_Tree{
private:struct segment_tree{int l,r;bool pd;char mn,mx;#define l(x) tree[x].l#define r(x) tree[x].r#define pd(x) tree[x].pd#define mn(x) tree[x].mn#define mx(x) tree[x].mx}tree[N<<2];
public:inline void pushup(int k){pd(k) = ((s[r(k<<1)] <= s[l(k<<1|1)]) && pd(k<<1) && pd(k<<1|1));mn(k) = min(mn(k<<1),mn(k<<1|1));mx(k) = max(mx(k<<1),mx(k<<1|1));}void build(int k,int l,int r){l(k) = l,r(k) = r;if(l == r) return pd(k) = true,mn(k) = mx(k) = s[l],void();int mid = (l+r)>>1;build(k<<1,l,mid);build(k<<1|1,mid+1,r);pushup(k);// cerr<<l(k)<<' '<<r(k)<<' '<<r(k<<1)<<' ' <<s[r(k<<1)]<<' '<<l(k<<1|1)<<' '<<s[l(k<<1|1)]<<'\n';}void update(int k,int pos){if(l(k) == r(k)) return mn(k) = mx(k) = s[pos],void();int mid = (l(k) + r(k))>>1;if(pos <= mid) update(k<<1,pos);else update(k<<1|1,pos);pushup(k);}bool query(int k,int l,int r){if(l <= l(k) && r(k) <= r) return pd(k);int mid = (l(k) + r(k)) >> 1;if(l > mid) return query(k<<1|1,l,r);if(r <= mid) return query(k<<1,l,r);return query(k<<1,l,r) && query(k<<1|1,l,r) && (s[r(k<<1)] <= s[l(k<<1|1)]);}char query_min(int k,int l,int r){if(l <= l(k) && r(k) <= r) return mn(k);int mid = (l(k) + r(k)) >> 1;if(r <= mid) return query_min(k<<1,l,r);if(l > mid) return query_min(k<<1|1,l,r);return min(query_min(k<<1,l,r),query_min(k<<1|1,l,r));}char query_max(int k,int l,int r){if(l <= l(k) && r(k) <= r) return mx(k);int mid = (l(k) + r(k)) >> 1;if(r <= mid) return query_max(k<<1,l,r);if(l > mid) return query_max(k<<1|1,l,r);return max(query_max(k<<1,l,r),query_max(k<<1|1,l,r));}
}St;
signed main(){cin.tie(nullptr)->sync_with_stdio(false);cout.tie(nullptr)->sync_with_stdio(false);cin>>n>>(s+1);for(int i = 0;i <= 26; ++i) T[i].mx = n;for(int i = 1;i <= n; ++i) T[s[i]-'a'].update(i,1);St.build(1,1,n);cin>>m;while(m--){int op,l,r,x;char c;cin>>op;if(op == 1){cin>>x>>c;T[s[x]-'a'].update(x,-1);s[x] = c;T[s[x]-'a'].update(x,1);St.update(1,x);}else{cin>>l>>r;if(!St.query(1,l,r)) cout<<"No\n";else{bool flag = true;int bg = St.query_min(1,l,r)-'a';int ed = St.query_max(1,l,r)-'a';for(int i = bg;i <= ed; ++i){if(i == s[l]-'a' || i == s[r] -'a') continue;if(l > 1?T[i].query(r) - T[i].query(l-1) != T[i].query(n):T[i].query(r) != T[i].query(n)){cout<<"No\n";flag = false;break;}}if(flag) cout<<"Yes\n";}}}
}

T3 精准打击

[ABC290G] Edge Elimination

从根节点考虑，若任意一棵子树都大于x，那么就将根节点的所有连边剪断，递归至任意子树。

如果任意一颗子树不大于x，那么就保留根节点，看看要保留几颗子树。如果有不满一颗的，递归处理。

注意根节点上是否要剪边。

就，没了

点此查看代码

#include<bits/stdc++.h>
#include<bits/extc++.h>
// using namespace __gnu_pbds;
// using namespace __gnu_cxx;
using namespace std;
#define infile(x) freopen(x,"r",stdin)
#define outfile(x) freopen(x,"w",stdout)
#define errfile(x) freopen(x,"w",stderr)
using ll=long long;using ull=unsigned long long;using db = double;
#ifdef LOCALFILE *InFile = infile("in.in"),*OutFile = outfile("out.out");// FILE *ErrFile=errfile("err.err");
#elseFILE *Infile = stdin,*OutFile = stdout;//FILE *ErrFile = stderr;
#endif
ll d,k,x;
signed main(){cin.tie(nullptr)->sync_with_stdio(false);cout.tie(nullptr)->sync_with_stdio(false);int T;cin>>T;while(T--){cin>>d>>k>>x;ll res = k;ll a[100];a[0] = 1;for(int i = 1;i <= d; ++i) a[i] = a[i-1] * k + 1;ll ans = 1e18;for(int i = 0;i <= d; ++i){if(a[i] >= x){ll res = !(i == d);ll tot = a[i] - x,j = i - 1;while(tot && j >= 0){res += tot/a[j];tot %= a[j];j--;}ans = min(ans,res);}}cout<<ans<<'\n';}
}

T4 你画我猜

[BJOI2018] 双人猜数游戏

两个神仙玩游戏，Alice不知道，Bob也不知道，几个不知道后，就都知道了。

太难了。跳了。