Problem 1. Barn Tree

Problem 2. Circular Barn

Problem 3. Range Reconstruction

Problem 1. Barn Tree

Farmer John's farm has NN barns (2≤N≤2⋅1052≤N≤2⋅105) numbered 1…N1…N. There are N−1N−1 roads, where each road connects two barns and it is possible to get from any barn to any other barn via some sequence of roads. Currently, the jjth barn has hjhj hay bales (1≤hj≤1091≤hj≤109).

To please his cows, Farmer John would like to move the hay such that each barn has an equal number of bales. He can select any pair of barns connected by a road and order his farmhands to move any positive integer number of bales less than or equal to the number of bales currently at the first barn from the first barn to the second.

Please determine a sequence of orders Farmer John can issue to complete the task in the minimum possible number of orders. It is guaranteed that a sequence of orders exists.

INPUT FORMAT (input arrives from the terminal / stdin):

The first line of input contains the value of N.N.

The second line of input contains the space-separated values of hjhj for j=1…Nj=1…N.

The final N−1N−1 lines of input each contain two space-separated barn numbers ui viui vi, indicating that there is a bidirectional road connecting uiui and vivi.

OUTPUT FORMAT (print output to the terminal / stdout):

Output the minimum possible number of orders, followed a sequence of orders of that length, one per line.

Each order should be formatted as three space-separated positive integers: the source barn, the destination barn, and the third describes the number of hay bales to move from the source to the destination.

If there are multiple solutions, output any.

SAMPLE INPUT:

SAMPLE OUTPUT:

In this example, there are a total of twelve hay bales and four barns, meaning each barn must have three hay bales at the end. The sequence of orders in the sample output can be verbally translated as below:

From barn 33 to barn 22, move 11 bale.
From barn 44 to barn 22, move 22 bales.
From barn 22 to barn 11, move 11 bale.

SCORING:

Test cases 2-8 satisfy N≤5000N≤5000
Test cases 7-10 satisfy vi=ui+1vi=ui+1
Test cases 11-16 satisfy no additional constraints

#include<bits/stdc++.h>
using namespace std;
#define mpp(a,b,c) make_pair(a,make_pair(b,c))
#define _1 first
#define _2 second.first
#define _3 second.second
#define pb push_back
#define eb emplace_back
#define int long long
namespace IO{inline int read(){int ans=0,flag=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')flag=-1;  ch=getchar();}while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();return ans*flag;}inline string reads(){string ans="";char ch=getchar();while(ch==' '||ch=='\n')ch=getchar();while(ch!=' '&&ch!='\n')ans+=ch,ch=getchar();  return ans;}inline char readc(){char ch=getchar();while(ch==' '||ch=='\n')ch=getchar();return ch;}inline int sqr(int x){return x*x;}inline void reada(int *a,int len){for(int i=1;i<=len;i++)a[i]=read();}  void write(int x){if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+'0');return;}
}
using IO::read;
const int N=4e5+7;
bool mem1=0,mem2=0;
int Datas=0,T=1,n,avg=0,h[N];
vector<int>g[N];
long long ned[N];
vector<pair<int,pair<int,long long> > >opt;
void solve(int u,int fa){for(auto v:g[u]){if(v==fa||ned[v]<=0)continue;opt.eb(mpp(u,v,ned[v]));solve(v,u); }
}
void dfs(int u,int fa){ for(auto v:g[u]){if(v==fa)continue;dfs(v,u); }for(auto v:g[u]){if(v==fa)continue;ned[u]+=ned[v];}ned[u]+=avg-h[u];if(ned[u]<=0){solve(u,fa);if(ned[u]!=0) opt.eb(mpp(u,fa,-ned[u]));}
}
void Main(){n=read();long long sum=0;for(int i=1;i<=n;i++)h[i]=read(),sum+=h[i];avg=sum/n;for(int i=1;i<n;i++){int u=read(),v=read(); g[u].pb(v),g[v].pb(u);}dfs(1,0);printf("%lld\n",(int)(opt.size()));for(auto x:opt){printf("%lld %lld %lld\n",x._1,x._2,x._3);}return;
}
signed main(){
#ifdef EXODUSprintf("%.5lfMB\n",abs(&mem2-&mem1)/1024.0/1024.0);freopen("Data.in","r",stdin);freopen("Code.out","w",stdout);
#endif
#ifdef online_judgefreopen("input.in","r",stdin);freopen("user_out.out","w",stdout);
#endifif(Datas)T=read();while(T--)Main();
#ifdef EXODUScerr<<clock()<<"ms"<<endl;
#endifreturn 0;
}

Problem 2. Circular Barn

Farmer John and his archnemesis Farmer Nhoj are playing a game in a circular barn. There are NN (1≤N≤1051≤N≤105) rooms in the barn, and the iith room initially contains aiai cows (1≤ai≤5⋅1061≤ai≤5⋅106). The game is played as follows:

Both farmers will always be in the same room. After entering a room, each farmer takes exactly one turn, with Farmer John going first. Both farmers initially enter room 11.
If there are zero cows in the current room, then the farmer to go loses. Otherwise, the farmer to go chooses an integer PP, where PP must either be 11 or a prime number at most the number of cows in the current room, and removes PP cows from the current room.
After both farmers have taken turns, both farmers move to the next room in the circular barn. That is, if the farmers are in room ii, then they move to room i+1i+1, unless they are in room NN, in which case they move to room 11.

Determine the farmer that wins the game if both farmers play optimally.

INPUT FORMAT (input arrives from the terminal / stdin):

The input contains TT test cases. The first line contains TT (1≤T≤10001≤T≤1000). Each of the TT test cases follow.

Each test case starts with a line containing NN, followed by a line containing a1,…,aNa1,…,aN.

It is guaranteed that the sum of all NN is at most 2⋅1052⋅105.

OUTPUT FORMAT (print output to the terminal / stdout):

For each test case, output the farmer that wins the game, either "Farmer John" or "Farmer Nhoj."

SAMPLE INPUT:

SAMPLE OUTPUT:

Farmer Nhoj
Farmer John
Farmer John
Farmer John
Farmer Nhoj

For the first test case, Farmer John can remove 11, 22, or 33 cows from the first room. Whichever number he removes, Nhoj can remove the remaining cow(s), forcing FJ to lose when they circle back to the first room.

For the second test case, FJ can remove 55 cows, forcing Nhoj to work with only 44 cows remaining. Now, Nhoj can either remove 11, 22, or 33 cows. This is now similar to the first test case.

For the third and fourth test cases, FJ can immediately remove all the cows from the first room, forcing Nhoj to lose.

For the fifth test case, FJ can remove 11, 22, or 33, cows from the first room, and Nhoj can remove the rest right after. When they circle back around to the first room, FJ will lose.

SCORING:

Inputs 2-4 satisfy N=1N=1.
Inputs 1, 2, and 5-7 satisfy ai≤1000ai≤1000.
Inputs 8-20 satisfy no additional constraints.

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define np pair<int,int>
#define mp(a,b) make_pair(a,b)
namespace IO{il int read(){int ans=0,flag=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')flag=-1;ch=getchar();}while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();return ans*flag;}il string reads(){string ans="";char ch=getchar();while(ch==' '||ch=='\n')ch=getchar();while(ch!=' '&&ch!='\n')ans+=ch,ch=getchar();return ans;}il char readc(){char ch=getchar();while(ch==' '||ch=='\n')ch=getchar();return ch;}il int sqr(int x){return x*x;}il void reada(int *a,int len){for(int i=1;i<=len;i++)a[i]=read();}void write(int x){if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+'0');return;}
}
using IO::read;
bool mem1=0;
int Datas=1,T=1;
const int N=5e6+7,P=5e6+7;
int pri[P],cnt=0,n,a[N];
bool chk[N];
void get_prime(int Up){chk[1]=0;pri[++cnt]=1;for(int i=2;i<=Up;i++){if(!chk[i])pri[++cnt]=i;for(int j=2;j<=cnt&&pri[j]*i<=Up;j++){chk[i*pri[j]]=1;if(i%pri[j]==0)break;}}
}
int res[N];
bool mem2=0;
void Main(){n=read();for(int i=1;i<=n;i++){a[i]=read();}np tim1=mp(2e9,2e9),tim2=mp(2e9,2e9);for(int i=1;i<=n;i++){if(a[i]%4==0){tim1=min(tim1,mp(a[i]/2/2+1,i));}else{if(a[i]&1^1)tim2=min(tim2,mp((a[i]/2+1)/2,i));else{ if(res[a[i]])tim2=min(tim2,mp((res[a[i]]+1)/2,i));else{for(int j=0;j<=a[i];j+=4){if(!chk[a[i]-j]){res[a[i]]=j/2+1; break; }}tim2=min(tim2,mp((res[a[i]]+1)/2,i));}}}}if(tim1<tim2)puts("Farmer Nhoj");else puts("Farmer John");return;
}
int main(){get_prime(5e6);
#ifdef EXODUSprintf("%.5lfMB\n",abs(&mem2-&mem1)/1024.0/1024.0);freopen("Data.in","r",stdin);freopen("Code.out","w",stdout);
#endif
#ifdef online_judgefreopen("input.in","r",stdin);freopen("user_out.out","w",stdout);
#endifif(Datas)T=read();while(T--)Main();
#ifdef EXODUScerr<<clock()<<"ms"<<endl;
#endifreturn 0;
}

Problem 3. Range Reconstruction

Bessie has an array a1,…,aNa1,…,aN, where 1≤N≤3001≤N≤300 and 0≤ai≤1090≤ai≤109 for all ii. She won't tell you aa itself, but she will tell you the range of each subarray of aa. That is, for each pair of indices i≤ji≤j, Bessie tells you ri,j=maxa[i…j]−mina[i…j]ri,j=maxa[i…j]−mina[i…j]. Given these values of rr, please construct an array that could have been Bessie's original array. The values in your array should be in the range [−109,109][−109,109].

INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains NN.

Another NN lines follow. The iith of these lines contains the integers ri,i,ri,i+1,…,ri,Nri,i,ri,i+1,…,ri,N.

It is guaranteed that there is some array aa with values in the range [0,109][0,109] such that for all i≤ji≤j, ri,j=maxa[i…j]−mina[i…j]ri,j=maxa[i…j]−mina[i…j].

OUTPUT FORMAT (print output to the terminal / stdout):

Output one line containing NN integers b1,b2,…,bNb1,b2,…,bN in the range [−109,109][−109,109] representing your array. They must satisfy ri,j=maxb[i…j]−minb[i…j]ri,j=maxb[i…j]−minb[i…j] for all i≤ji≤j.

SAMPLE INPUT:

SAMPLE OUTPUT:

1 3 2

For example, r1,3=maxa[1…3]−mina[1…3]=3−1=2r1,3=maxa[1…3]−mina[1…3]=3−1=2.

SAMPLE INPUT:

SAMPLE OUTPUT:

0 1 1

This example satisfies the constraints for subtask 1.

SAMPLE INPUT:

SAMPLE OUTPUT:

1 2 3 2

This example satisfies the constraints for subtask 2.

SAMPLE INPUT:

SAMPLE OUTPUT:

1 2 2 0

SCORING:

Test 5 satisfies r1,N≤1r1,N≤1.
Tests 6-8 satisfy ri,i+1=1ri,i+1=1 for all 1≤i<N1≤i<N.
Tests 9-14 satisfy no additional constraints.

#include<bits/stdc++.h>
using namespace std;
const int inf=0x3f3f3f3f;
const int Maxn=310;
int n,a[Maxn][Maxn],ans[Maxn],minn[Maxn],maxn[Maxn];
bool vis[Maxn];
int main(){scanf("%d",&n);for(int i=1;i<=n;i++)for(int j=i;j<=n;j++) scanf("%d",&a[i][j]); int pos=n;ans[pos]=0;int minn=0,maxx=0;memset(vis,-1,sizeof(vis));pos--;while(pos!=0){int ans1=ans[pos+1]+a[pos][pos+1],ans2=ans[pos+1]-a[pos][pos+1]; int minn1=ans1,maxn1=ans1;int minn2=ans2,maxn2=ans2; bool flag1=1,flag2=1;for(int i=pos+1;i<=n;i++){minn1=min(minn1,ans[i]);minn2=min(minn2,ans[i]);maxn1=max(maxn1,ans[i]);maxn2=max(maxn2,ans[i]);if(maxn2-minn2!=a[pos][i]) flag2=0;  if(maxn1-minn1!=a[pos][i]) flag1=0; }if(!flag1 && !flag2){  pos++;vis[pos]=-1;continue;}if(flag1 && !flag2){ans[pos]=ans1;pos--;}else if(flag2 && !flag1){ans[pos]=ans2;pos--;}else if(flag1 && flag2){if(vis[pos]==-1){ans[pos]=ans1;vis[pos]=1;pos--;}else{ans[pos]=ans2;vis[pos]=2;pos--;}}}int awa=inf;for(int i=1;i<=n;i++)awa=min(awa,ans[i]); for(int i=1;i<n;i++)printf("%d ",ans[i]-awa); printf("%d\n",ans[n]-awa);return 0;
}

USACO 12月 2022-2023 December Contest Silver银组题解相关推荐

USACO 12月 2020-2021 December Contest Silver银组题解
Problem 1. Cowntagion Farmer John and his fellow farmers have been working nonstop to control the sp ...
USACO 1月 2021-2022 January Contest Silver银组题解
你好啊我又又又来了要准备usaco的铁铁们可以参考这个文章哦! 想刷好USACO--看这篇文章就够了_GeekAlice的博客-CSDN博客我最近是发现了一个很好用的网站https://blog.c ...
2022年蓝桥杯C++B组题解 - 很详细
本人这次侥幸省1,特做题解复习,哈哈哈- 1.进制转换(5分): 问题描述: 直接计算 2 + 2 * 9 + 2 * 9 * 9 * 9 答案: 1478 2.顺子日期(5分) 这题有争议: 主要在 ...
leaf优秀作品推荐天使のいない12月（没有天使的12月）天使不在的12月攻略
推荐理由: 不愧是leaf的作品,人设以及CG的确很美这个恐怕是LEAF第一个能让大师有爱的作品,像什么ToHeart 2 AD ,玩的时候差点就睡着了.... 如果是抱着玩童话式的恋爱游戏的心态 ...
前女幽魂服务器维护,【维护】2020年12月10日在线维护公告（新区：千年调-相见欢）...
本周四在线更新内容如下: 系统 1.好友召回系统迭代全服投放.迭代后可召回更多流失好友,流失好友回流后可与召回玩家进行绑定,共同完成任务领取丰厚奖励. 玩法 1.第七届跨服明星赛总决赛将于12月1 ...
2022 12月——2023 2月寒假假期总结
在寒假期间,本人有效的贯彻落实了<科研小分队寒假计划>.做到了以数据结构学习为主体,充分利用考试题.算法题.力扣题一体化的学习优势,圆满完成了本人所学习的数据结构等目的.完成了科研小分队寒 ...
【记录】ChatGPT｜注册流程、使用技巧与应用推荐（更新至2022年12月14日）
昨天,2022年12月13日,在下午和晚上,ChatGPT 就开始因为请求过多而写到一半就崩溃,出现network error,可见它的关注度确实是越来越可观了. 正好最近世界杯,有博客活动, ...
基因编辑相关最新研究进展（2022年12月）
[1]西湖大学马丽佳团队开发新型CRISPR脱靶和DNA易位检测工具 2022-12-15报道,2022年12月12日,西湖大学生命科学学院马丽佳团队在 Nature Communications 期 ...
2022年12月国产数据库大事记-墨天轮
本文为墨天轮技术社区整理的2022年12月国产数据库大事件和重要产品发布消息. 目录 12月国产数据库大事记(时间线) 产品/版本发布兼容认证排行榜新增数据库厂商活动相关资料 12月国产数据库 ...

USACO 12月 2022-2023 December Contest Silver银组题解

Problem 1. Barn Tree

Problem 2. Circular Barn

Problem 3. Range Reconstruction

USACO 12月 2022-2023 December Contest Silver银组题解相关推荐

最新文章

热门文章

USACO 12月 2022-2023 December Contest Silver银组 题解

Problem 1. Barn Tree

Problem 2. Circular Barn

Problem 3. Range Reconstruction

USACO 12月 2022-2023 December Contest Silver银组 题解相关推荐

最新文章

热门文章

USACO 12月 2022-2023 December Contest Silver银组题解

USACO 12月 2022-2023 December Contest Silver银组题解相关推荐