2020-2021:teams:wangzai_milk:20200614比赛记录
| 两侧同时换到之前的修订记录 前一修订版 后一修订版 | 前一修订版 | ||
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2020-2021:teams:wangzai_milk:20200614比赛记录 [2020/06/21 23:06] infinity37 [比赛总结] |
2020-2021:teams:wangzai_milk:20200614比赛记录 [2020/06/25 21:21] (当前版本) wzx27 |
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| ===== 题解 ===== | ===== 题解 ===== | ||
| - | ===== replay ===== | + | ==== A.Advertising Strategy ==== |
| - | ===== 比赛总结 ===== | + | 目标订购者为$n$，总花费$k$。$a_i$为当前订购者数量，每天可以花费$x_i$使得$b_i=a_i+x_i$，之后$a_{i+1}=b_i+\min(b_i,\lfloor\frac{n-b_i}2\rfloor)$，问达到目标订购的最少天数 |
| + | 题解：显然至少要有花费要在最后一天给出，其他钱应该是早花早受益，于是在第一天和最后一天花钱。枚举在第一天花的钱，模拟天数增加直到剩下的人可以一次性花钱补全。 | ||
| + | |||
| + | code: | ||
| + | |||
| + | <hidden> | ||
| + | <code cpp> | ||
| + | #include<bits/stdc++.h> | ||
| + | #define ll long long | ||
| + | #define pii_ pair<int,int> | ||
| + | #define mp_ make_pair | ||
| + | #define pb push_back | ||
| + | #define fi first | ||
| + | #define se second | ||
| + | #define rep(i,a,b) for(int i=(a);i<=(b);i++) | ||
| + | #define per(i,a,b) for(int i=(a);i>=(b);i--) | ||
| + | #define show1(a) cout<<#a<<" = "<<a<<endl | ||
| + | #define show2(a,b) cout<<#a<<" = "<<a<<"; "<<#b<<" = "<<b<<endl | ||
| + | using namespace std; | ||
| + | const ll INF = 1LL<<60; | ||
| + | const int inf = 1<<30; | ||
| + | const int maxn = 2e5+5; | ||
| + | inline void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);} | ||
| + | |||
| + | ll n,k; | ||
| + | |||
| + | int main() | ||
| + | { | ||
| + | fastio(); | ||
| + | cin>>n>>k; | ||
| + | ll res=INF; | ||
| + | for(ll i=1;i<k;i++) | ||
| + | { | ||
| + | ll now = k-i,ans = 1; | ||
| + | while(now<n-i){ | ||
| + | now = now + min(now,(n-now)/2); | ||
| + | ans++; | ||
| + | } | ||
| + | res = min(res,ans); | ||
| + | } | ||
| + | cout<<res<<endl; | ||
| + | return 0; | ||
| + | } | ||
| + | </code> | ||
| + | </hidden> | ||
| + | \\ | ||
| + | |||
| + | ==== C.Carpet ==== | ||
| + | |||
| + | 要将$n(1\le n\le 100000)$个点的树铺在$1000000×20$的坐标平面里，使得边与边无端点之外的重合点。 | ||
| + | |||
| + | 题解：树剖，将轻儿子往上放，重儿子往右放。 | ||
| + | |||
| + | code: | ||
| + | |||
| + | <hidden> | ||
| + | <code cpp> | ||
| + | #include<bits/stdc++.h> | ||
| + | using namespace std; | ||
| + | const int N=1e5+5; | ||
| + | int n,head[N],cnt,sz[N]; | ||
| + | int x[N],y[N],tot[25]; | ||
| + | struct Node | ||
| + | { | ||
| + | int nxt,to; | ||
| + | }Edges[N*2]; | ||
| + | void addedge(int u,int v) | ||
| + | { | ||
| + | Edges[++cnt].nxt=head[u]; | ||
| + | head[u]=cnt; | ||
| + | Edges[cnt].to=v; | ||
| + | } | ||
| + | void dfs1(int u,int f) | ||
| + | { | ||
| + | sz[u]=1; | ||
| + | for(int i=head[u];~i;i=Edges[i].nxt) | ||
| + | { | ||
| + | int v=Edges[i].to; | ||
| + | if(v==f)continue; | ||
| + | dfs1(v,u),sz[u]+=sz[v]; | ||
| + | } | ||
| + | } | ||
| + | void dfs2(int u,int f,int d) | ||
| + | { | ||
| + | y[u]=d,x[u]=++tot[d]; | ||
| + | int maxv=0,k=0; | ||
| + | for(int i=head[u];~i;i=Edges[i].nxt) | ||
| + | { | ||
| + | int v=Edges[i].to; | ||
| + | if(v==f)continue; | ||
| + | if(sz[v]>maxv)maxv=sz[v],k=v; | ||
| + | } | ||
| + | for(int i=head[u];~i;i=Edges[i].nxt) | ||
| + | { | ||
| + | int v=Edges[i].to; | ||
| + | if(v==f||v==k)continue; | ||
| + | dfs2(v,u,d+1); | ||
| + | } | ||
| + | if(k)dfs2(k,u,d); | ||
| + | } | ||
| + | int main() | ||
| + | { | ||
| + | memset(head,-1,sizeof(head)); | ||
| + | scanf("%d",&n); | ||
| + | for(int i=1;i<n;i++) | ||
| + | { | ||
| + | int u,v; | ||
| + | scanf("%d%d",&u,&v); | ||
| + | addedge(u,v),addedge(v,u); | ||
| + | } | ||
| + | dfs1(1,0),dfs2(1,0,1); | ||
| + | for(int i=1;i<=n;i++)printf("%d %d\n",x[i],y[i]); | ||
| + | return 0; | ||
| + | } | ||
| + | </code> | ||
| + | </hidden> | ||
| ==== D. Decoding of Varints ==== | ==== D. Decoding of Varints ==== | ||
| === 题意 === | === 题意 === | ||
| - | 语文题，就是给了一个类似128进制的定义，然后给一个未知的序列，对于这个序列中的每一个数字 | + | 语文题，就是给了一个类似128进制的定义，然后给一个未知的序列，对于这个序列中的每一个数字，如果这个数字$\geq 0$的时候，这个数就变为这个数字的二倍，奇遇的时候是这个数字相反数的二倍再减一，把新数字搞成他之前定义的形式中的每一位，现在给你每一位，让你装换回去。 |
| + | |||
| + | === 数据范围 === | ||
| + | $n\leq 10000$ | ||
| + | |||
| + | === 题解 === | ||
| + | |||
| + | 就 模拟啊 | ||
| + | |||
| + | === 代码 === | ||
| + | <hidden> | ||
| + | <code c++> | ||
| + | #include <stdio.h> | ||
| + | #include <string.h> | ||
| + | #include <stdlib.h> | ||
| + | #include <algorithm> | ||
| + | using namespace std; | ||
| + | const int N = 1e4+5; | ||
| + | int a[N]; | ||
| + | typedef unsigned long long ll; | ||
| + | int main() | ||
| + | { | ||
| + | int n; | ||
| + | scanf("%d",&n); | ||
| + | for (int i = 1;i<= n;i++)scanf("%d",&a[i]); | ||
| + | for (int i = 1;i<= n;i++) { | ||
| + | int y = i; | ||
| + | ll ans = 0,x = 1; | ||
| + | while (y <= n) { | ||
| + | if (a[y]>=128) { | ||
| + | ans += (a[y]-128)*x; | ||
| + | x *= 128; | ||
| + | y++; | ||
| + | } else { | ||
| + | ans += a[y]*x; | ||
| + | i = y; | ||
| + | break; | ||
| + | } | ||
| + | } | ||
| + | if (ans & 1) | ||
| + | printf("%lld\n",(long long)0-((ans-1)/2+1)); | ||
| + | else | ||
| + | printf("%llu\n",ans/2); | ||
| + | } | ||
| + | return 0; | ||
| + | } | ||
| + | </code> | ||
| + | </hidden> | ||
| + | \\ | ||
| + | |||
| + | |||
| + | ==== F. Fake or Leak ==== | ||
| + | 反正就是大模拟。。码力和英语都好差，写了好多次才过 | ||
| + | |||
| + | <hidden> | ||
| + | <code cpp> | ||
| + | #include<bits/stdc++.h> | ||
| + | #define ll long long | ||
| + | #define pii_ pair<int,int> | ||
| + | #define mp_ make_pair | ||
| + | #define pb push_back | ||
| + | #define fi first | ||
| + | #define se second | ||
| + | #define rep(i,a,b) for(int i=(a);i<=(b);i++) | ||
| + | #define per(i,a,b) for(int i=(a);i>=(b);i--) | ||
| + | #define show1(a) cout<<#a<<" = "<<a<<endl | ||
| + | #define show2(a,b) cout<<#a<<" = "<<a<<"; "<<#b<<" = "<<b<<endl | ||
| + | using namespace std; | ||
| + | const ll INF = 1LL<<60; | ||
| + | const int inf = 1<<30; | ||
| + | const int maxn = 1005; | ||
| + | inline void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);} | ||
| + | |||
| + | int n,m,k,a,t; | ||
| + | int nowTime[maxn],nowSolve[maxn],has[maxn],bestTime[maxn],in[maxn],finalAc[maxn]; | ||
| + | char op; | ||
| + | string name,laterName[maxn],rn[maxn]; | ||
| + | map<string,int> mp; | ||
| + | |||
| + | void GG() | ||
| + | { | ||
| + | cout<<"Fake"<<endl; | ||
| + | exit(0); | ||
| + | } | ||
| + | struct node | ||
| + | { | ||
| + | string name; | ||
| + | int solve,time,fac; | ||
| + | bool operator <(node e){ | ||
| + | if(solve!=e.solve) return solve>e.solve; | ||
| + | if(time!=e.time) return time<e.time; | ||
| + | if(fac!=e.fac) return fac<e.fac; | ||
| + | return name<e.name; | ||
| + | } | ||
| + | }; | ||
| + | vector<node> team; | ||
| + | vector<string> res; | ||
| + | int main() | ||
| + | { | ||
| + | fastio(); | ||
| + | cin>>n>>m>>k; | ||
| + | rep(j,1,m){ | ||
| + | cin>>name; | ||
| + | mp[name] = j; | ||
| + | rn[j] = name; | ||
| + | rep(i,1,n){ | ||
| + | cin>>op>>a>>t; | ||
| + | if(op=='+'){ | ||
| + | nowTime[j] += 20*(a-1)+t; | ||
| + | nowSolve[j]++; | ||
| + | bestTime[j] += 20*(a-1)+t; | ||
| + | finalAc[j] = max(finalAc[j],t); | ||
| + | } | ||
| + | else bestTime[j] += 20*a+240; | ||
| + | } | ||
| + | } | ||
| + | rep(j,1,k){ | ||
| + | cin>>laterName[j]; | ||
| + | name = laterName[j]; | ||
| + | int ti = 0,so = 0,fac=0; | ||
| + | has[mp[name]] = 1; | ||
| + | rep(i,1,n){ | ||
| + | cin>>op>>a>>t; | ||
| + | if(op=='+'){ | ||
| + | ti += 20*(a-1)+t; | ||
| + | so++; | ||
| + | fac = max(fac,t); | ||
| + | } | ||
| + | } | ||
| + | team.pb((node){name,so,ti,fac}); | ||
| + | } | ||
| + | rep(i,1,m){ | ||
| + | if(!has[i]){ | ||
| + | team.pb((node){rn[i],nowSolve[i],nowTime[i],finalAc[i]}); | ||
| + | if(nowSolve[i]<n) team.pb((node){rn[i],n,bestTime[i],240}); | ||
| + | else team.pb((node){rn[i],n,bestTime[i],finalAc[i]}); | ||
| + | } | ||
| + | } | ||
| + | sort(team.begin(),team.end()); | ||
| + | int cnt = 0,flag = 0; | ||
| + | for(node x:team){ | ||
| + | if(has[mp[x.name]]) flag = 1; | ||
| + | if(flag && cnt<k){ | ||
| + | if(has[mp[x.name]]) {res.pb(x.name);cnt++;} | ||
| + | else { | ||
| + | in[mp[x.name]]++; | ||
| + | if(in[mp[x.name]]==2) GG(); | ||
| + | } | ||
| + | } | ||
| + | } | ||
| + | rep(i,1,k){ | ||
| + | if(res[i-1]!=laterName[i]) GG(); | ||
| + | } | ||
| + | cout<<"Leaked"<<endl; | ||
| + | return 0; | ||
| + | } | ||
| + | |||
| + | </code> | ||
| + | </hidden> | ||
| + | |||
| + | \\ | ||
| + | |||
| + | ==== G. God of Winds ==== | ||
| + | 一个网格图，每个小正方形的边有个权值。可以任选格子，加入一个顺时针或者逆时针的风，用来改变格子周围四条边的权值。右和下是正方形，题目给定权值，问这样的图是否存在。 | ||
| + | |||
| + | 注意到每条边只有两个格子能影响它，假设每个 $(i,j)$ 处的格子有 $x_{i,j}$ 个顺时针方向的风，那么整个图就是个等式的差分约束，对每个变量连边然后 $\text{dfs}$ 一遍，判断会不会有矛盾的点。 | ||
| + | |||
| + | <hidden> | ||
| + | <code cpp> | ||
| + | #include<bits/stdc++.h> | ||
| + | #define ll long long | ||
| + | #define pii_ pair<int,int> | ||
| + | #define mp_ make_pair | ||
| + | #define pb push_back | ||
| + | #define fi first | ||
| + | #define se second | ||
| + | #define rep(i,a,b) for(int i=(a);i<=(b);i++) | ||
| + | #define per(i,a,b) for(int i=(a);i>=(b);i--) | ||
| + | #define show1(a) cout<<#a<<" = "<<a<<endl | ||
| + | #define show2(a,b) cout<<#a<<" = "<<a<<"; "<<#b<<" = "<<b<<endl | ||
| + | using namespace std; | ||
| + | const ll INF = 1LL<<60; | ||
| + | const int inf = 1<<30; | ||
| + | inline void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);} | ||
| + | const int maxn = 250005; | ||
| + | int n,m; | ||
| + | int r[505][505],c[505][505]; | ||
| + | int vis[maxn],head[maxn],tot; | ||
| + | ll dis[maxn]; | ||
| + | struct edge | ||
| + | { | ||
| + | int u,v,w,nxt; | ||
| + | }es[maxn<<2]; | ||
| + | void add(int u,int v,int w) | ||
| + | { | ||
| + | es[++tot] = (edge){u,v,w,head[u]}; | ||
| + | head[u] = tot; | ||
| + | } | ||
| + | void dfs(int u) | ||
| + | { | ||
| + | vis[u] = 1; | ||
| + | for(int i=head[u];~i;i=es[i].nxt){ | ||
| + | int v = es[i].v,w = es[i].w; | ||
| + | if(!vis[v]){ | ||
| + | dis[v] = dis[u] + w; | ||
| + | dfs(v); | ||
| + | }else{ | ||
| + | if(dis[v]!=dis[u]+w){ | ||
| + | cout<<"No"<<endl; | ||
| + | exit(0); | ||
| + | } | ||
| + | } | ||
| + | } | ||
| + | } | ||
| + | int main() | ||
| + | { | ||
| + | fastio(); | ||
| + | memset(head,-1,sizeof(head)); | ||
| + | cin>>n>>m; | ||
| + | rep(i,0,n-1)rep(j,0,m-1){ | ||
| + | cin>>r[i][j]>>c[i][j]; | ||
| + | } | ||
| + | rep(i,0,n-1){ | ||
| + | rep(j,0,m-1){ | ||
| + | int u = i*m+j,v = (i-1+n)%n*m + j; | ||
| + | add(u,v,r[i][j]);add(v,u,-r[i][j]); | ||
| + | int x = i*m+j,y = i*m + (j-1+m)%m; | ||
| + | add(x,y,-c[i][j]);add(y,x,c[i][j]); | ||
| + | } | ||
| + | } | ||
| + | dfs(0); | ||
| + | cout<<"Yes"<<endl; | ||
| + | return 0; | ||
| + | } | ||
| + | </code> | ||
| + | </hidden> | ||
| + | |||
| + | \\ | ||
| + | |||
| + | ==== H. Hilarious Cooking ==== | ||
| + | === 题意 === | ||
| + | 希望你构造一个序列，序列相邻两个数差值不超过1，和为T，然后其中有一些给定位置的给定数字，问能否构造成功。 | ||
| + | |||
| + | === 数据范围 === | ||
| + | 序列的长度$n\leq 2\times 10^9$，$1\leq T\leq 10^{18}$ | ||
| + | |||
| + | 给定的数字个数$1\leq m\leq 100000$ | ||
| + | |||
| + | === 题解 === | ||
| + | 可以考虑，构造这样一个序列可以保证在一个范围内连续，因为除非在最大值和最小值的位置，其他情况下都可以找到一个位置+1或者找到一个位置-1并符合序列规定，那么我们要做的就是确定这个序列可能的最大值和可能的最小值。注意到其限制其实就是给定的数字，最大值在两个给定的数字构造一个像山顶或山坡的形状，最小值就构造一个山谷或山坡一样的形状即可，要注意一些细节，比如最开始和最后的。 | ||
| + | |||
| + | === 代码 === | ||
| + | |||
| + | <hidden> | ||
| + | <code c++> | ||
| + | #include <stdio.h> | ||
| + | #include <string.h> | ||
| + | #include <stdlib.h> | ||
| + | #include <algorithm> | ||
| + | using namespace std; | ||
| + | typedef long long ll; | ||
| + | const int N = 1e5+5; | ||
| + | ll x[N],y[N]; | ||
| + | ll T,n; | ||
| + | ll calc(ll x,ll y) { | ||
| + | if (x > y || y < 0) return 0; | ||
| + | ll nx = x < 0 ? -x : 0; | ||
| + | return (x+y)*(y-x+1)/2 + (nx+1)*nx/2; | ||
| + | } | ||
| + | int main() | ||
| + | { | ||
| + | scanf("%lld",&T); | ||
| + | ll minn = 0; | ||
| + | ll maxn = 0; | ||
| + | int m; | ||
| + | scanf("%lld%d",&n,&m); | ||
| + | for (int i = 1;i<= m;i++) { | ||
| + | scanf("%lld%lld",&x[i],&y[i]); | ||
| + | minn += y[i]; | ||
| + | maxn += y[i]; | ||
| + | } | ||
| + | for (int i = 1;i < m;i++) { | ||
| + | ll dis = x[i+1]-x[i]; | ||
| + | ll val = abs(y[i+1]-y[i]); | ||
| + | if (val > dis) { | ||
| + | printf("No\n"); | ||
| + | return 0; | ||
| + | } | ||
| + | if (dis == 1) { continue; } | ||
| + | ll maxx = max(y[i+1],y[i]) + (dis-val)/2; | ||
| + | ll minx = min(y[i+1],y[i]) - (dis-val)/2; | ||
| + | if (val % 2 == dis % 2) { | ||
| + | maxn += calc(max(y[i+1],y[i]),maxx) + calc(min(y[i+1],y[i]),maxx) - max(maxx,0ll); | ||
| + | minn += calc(minx,max(y[i+1],y[i])) + calc(minx,min(y[i+1],y[i])) - max(minx,0ll); | ||
| + | |||
| + | } else { | ||
| + | maxn += calc(max(y[i+1],y[i]),maxx) + calc(min(y[i+1],y[i]),maxx); | ||
| + | minn += calc(minx,max(y[i+1],y[i])) + calc(minx,min(y[i+1],y[i])); | ||
| + | } | ||
| + | maxn -= y[i+1]+y[i]; | ||
| + | minn -= y[i+1]+y[i]; | ||
| + | } | ||
| + | maxn += calc(y[1]+1,y[1]+x[1]-1); | ||
| + | minn += calc(y[1]-x[1]+1,y[1]-1); | ||
| + | maxn += calc(y[m]+1,y[m]+n-x[m]); | ||
| + | minn += calc(y[m]-n+x[m],y[m]-1); | ||
| + | if (T<= maxn && T >= minn) { printf("Yes\n"); } | ||
| + | else { printf("No\n"); } | ||
| + | return 0; | ||
| + | } | ||
| + | </code> | ||
| + | </hidden> | ||
| + | \\ | ||
| + | ===== replay ===== | ||
| + | |||
| + | ===== 比赛总结 ===== | ||
| + | |||
2020-2021/teams/wangzai_milk/20200614比赛记录.1592751999.txt.gz · 最后更改: 2020/06/21 23:06 由 infinity37
