标签：rand absolute ror amp googl error Once visitor pre
https://codingcompetitions.withgoogle.com/kickstart/round/000000000019ff08/0000000000386d5c
（第一题ac，第二题ac，第三题，小数据集能ac，第四题，小数据集能ac）
第一题：
Isyana is given the number of visitors at her local theme park on N consecutive days. The number of visitors on the i-th day is Vi. A day is record breaking if it satisfies both of the following conditions:
Note that the very first day could be a record breaking day!
Please help Isyana find out the number of record breaking days.
The first line of the input gives the number of test cases, T. T test cases follow. Each test case begins with a line containing the integer N. The second line contains N integers. The i-th integer is Vi.
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the number of record breaking days.
Time limit: 20 seconds per test set.
Memory limit: 1GB.
1 ≤ T ≤ 100.
0 ≤ Vi ≤ 2 × 105.
1 ≤ N ≤ 1000.
1 ≤ N ≤ 2 × 105 for at most 10 test cases.
For the remaining cases, 1 ≤ N ≤ 1000.
Input |
Output |
4 8 1 2 0 7 2 0 2 0 6 4 8 15 16 23 42 9 3 1 4 1 5 9 2 6 5 6 9 9 9 9 9 9 |
Case #1: 2 Case #2: 1 Case #3: 3 Case #4: 0 |
#include<bits/stdc++.h> using namespace std; int main() { int t, n; cin>>t; for(int i=1;i<=t;i++) { cin>>n; vector<int> v; for(int j=0;j<n;j++) { int tt; cin>>tt; v.push_back(tt); } int result = 0; int max_v = INT_MIN; if(v.size()==1) result = 1; else { for(int j=0;j<v.size();j++) { if(j>=1) max_v = max(max_v, v[j-1]); if(j<=v.size()-2) { if(v[j]>max_v && v[j]>v[j+1]) result++; } else { if(v[j]>max_v) result++; } } } printf("Case #%d: %d\n", i, result); }//end of for return 0; }
第二题：
An alien has just landed on Earth, and really likes our music. Lucky for us.
The alien would like to bring home its favorite human songs, but it only has a very strange instrument to do it with: a piano with just 4 keys of different pitches.
The alien converts a song by writing it down as a series of keys on the alien piano. Obviously, this piano will not be able to convert our songs completely, as our songs tend to have many more than 4 pitches.
The alien will settle for converting our songs with the following rules instead:
Note: two notes with the same pitch do not need to be converted into the same key if they are not adjacent.
What the alien wants to know is: how often will it have to break its rules when converting a particular song?
To elaborate, let us describe one of our songs as having K notes. The first note we describe as "note 1", the second note "note 2", and the last note "note K."
So note 2 comes immediately after note 1.
Now if note 2 is lower than note 1 in our version of the song, yet converted to an equally-pitched or lower-pitched key (relative to note 2‘s conversion) in the alien‘s version of the song, then we consider that a single rule break.
For each test case, return the minimum amount of times the alien must necessarily break one of its rules in converting that song.
The first line of the input gives the number of test cases, T. T test cases follow.
Each test case consists of two lines.
The first line consists of a single integer, K.
The second line consists of K space-separated integers, A1, A2 ... AK, where Ai refers to the pitch of the i-th note for this test case.
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the minimum number of times that particular test case will require the alien to break its own rules during the conversion process.
Memory limit: 1GB.
1 ≤ T ≤ 100.
1 ≤ Ai ≤ 106.
Time limit: 20 seconds.
1 ≤ K ≤ 10.
Time limit: 40 seconds.
1 ≤ K ≤ 104.
Input |
Output |
2 5 1 5 100 500 1 8 2 3 4 5 6 7 8 9 |
Case #1: 0 Case #2: 1 |
#include<bits/stdc++.h> using namespace std; int main() { int t, a; cin>>t; for(int i=1;i<=t;i++) { cin>>a; vector<int> v; for(int j=0;j<a;j++) { int tt; cin>>tt; v.push_back(tt); } int result = 0; int inc = 0, dec = 0; for(int j=1;j<v.size();j++) { if(v[j]>v[j-1]) { inc++; dec = 0; } if(v[j]<v[j-1]) { dec++; inc = 0; } if(inc==4 || dec==4) { result++; inc = 0; dec = 0; } } printf("Case #%d: %d\n", i, result); } return 0; }
第三题：
Amadea and Bilva are decorating a rooted tree containing N nodes, labelled from 1 to N. Node 1 is the root of the tree, and all other nodes have a node with a numerically smaller label as their parent.
Amadea and Bilva‘s decorate the tree as follows:
The beauty of the tree is equal to the number of nodes painted at least once by either Amadea or Bilva. Note that even if they both paint a node, it only counts once.
What is the expected beauty of the tree?
The first line of the input gives the number of test cases, T. T test cases follow. Each test case begins with a line containing the three integers N, A and B. The second line contains N-1 integers. The i-th integer is the parent of node i+1.
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the expected beauty of the tree.
y will be considered correct if it is within an absolute or relative error of 10-6 of the correct answer. See the FAQ for an explanation of what that means, and what formats of real numbers we accept.
Memory limit: 1GB.
1 ≤ T ≤ 100.
1 ≤ A ≤ N.
1 ≤ B ≤ N.
Time limit: 20 seconds.
1 ≤ N ≤ 100.
Time limit: 40 seconds.
For up to 5 cases, 1 ≤ N ≤ 5 × 105.
For all other cases, 1 ≤ N ≤ 100.
Input |
Output |
3 8 2 3 1 1 3 4 4 3 4 10 3 4 1 1 1 1 1 1 1 1 1 4 3 1 1 2 3 |
Case #1: 2.65625 Case #2: 1.9 Case #3: 2.875 |
#include<bits/stdc++.h> using namespace std; struct TreeNode { int val; //vector<TreeNode*> child; TreeNode *parent; TreeNode(int x) : val(x), parent(NULL){} }; int main() { int t; cin>>t; for(int i=1;i<=t;i++) { int n,a,b; cin>>n>>a>>b; vector<TreeNode*> v(n+1,NULL); v[1] = new TreeNode(1); for(int j=0;j<n-1;j++) { int tt; cin>>tt; v[j+2] = new TreeNode(j+2); v[j+2] -> parent = v[tt]; } //vector<TreeNode*> vv(v); set<TreeNode*> s; int num = 0; for(int x=1;x<=n;x++) for(int y=1;y<=n;y++) { s.clear(); TreeNode* p = v[x]; TreeNode* q = v[y]; while(p) { s.insert(p); for(int k=1;k<=a;k++) { if(!p) break; p = p->parent; } }// while(q) { s.insert(q); for(int k=1;k<=b;k++) { if(!q) break; q = q->parent; } } num += s.size(); }//end of for x,y //cout<<"num:"<<num<<endl; //cout<<"n:"<<n<<endl; double result = num/1.0/(n*n); //cout<<"result:"<<result<<endl; printf("Case #%d: %f\n", i, result); } return 0; }
第四题：
Bangles is preparing to go on a tour of her local museum. The museum is made up of N rooms in a row, numbered from 1 to N from left to right. The rooms are connected by N-1 locked doors, each connecting a pair of adjacent rooms. Each door has a difficulty level indicating how difficult it is for Bangles to open the door. No two doors will have the same difficulty level. The door between the i-th room and (i+1)-th room has difficulty level Di.
Bangles will pick one of the rooms to start in, and visit each of the rooms in the museum one at a time, taking pictures as she goes. She takes a picture in her starting room, then she repeats the following procedure until she has taken a picture in all the rooms: Of the two locked doors available to her, she will open the door with the lower difficulty level and take a picture in the newly unlocked room. If there is only one locked door available to her, then she will unlock that door. Once a door is unlocked, it remains unlocked.
Bangles is not yet sure which room she would like to start in, so she needs you to answer Q queries. For the i-th query, she would like to know: What is the Ki-th room that she will take a picture in if she starts in the Si-th room?
The first line of the input gives the number of test cases, T. T test cases follow. The first line of each test case contains the two integers N and Q. The second line contains N-1 integers, describing the locked doors. The i-th integer (starting from 1) is Di. Then, Q lines follow, describing the queries. The i-th of these lines contains the two integers Si and Ki.
For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is a list of the answers for the Q queries in order, separated by spaces.
Time limit: 40 seconds per test set.
Memory limit: 1GB.
1 ≤ T ≤ 100.
1 ≤ Di ≤ 105, for all i.
All Di are distinct.
1 ≤ Si ≤ N, for all i.
1 ≤ Ki ≤ N, for all i.
2 ≤ N ≤ 1000.
1 ≤ Q ≤ 1000.
2 ≤ N ≤ 105 and 1 ≤ Q ≤ 105 for at most 20 test cases.
For the remaining cases, 2 ≤ N ≤ 1000 and 1 ≤ Q ≤ 1000.
Input |
Output |
2 5 4 90 30 40 60 3 4 3 1 1 5 4 3 10 2 6 2 4 5 9 30 7 1 8 6 8 6 8 |
#include<bits/stdc++.h> using namespace std; int main() { int t; cin>>t; for(int i=1;i<=t;i++) { int n, q; cin>>n>>q; vector<int> v; for(int j=0;j<n-1;j++) { int tt; cin>>tt; v.push_back(tt); } vector<int> result; for(int tt=1;tt<=q;tt++) { //cout<<"tt:"<<tt<<endl; vector<int> l_room(n+1,0),r_room(n+1,0),l_value(n+1,0),r_value(n+1,0); for(int j=1;j<=n;j++) { //cout<<"j:"<<j<<endl; l_room[j] = j-1; r_room[j] = j+1; if(j-2>=0) l_value[j] = v[j-2]; else l_value[j] = INT_MAX; if(j!=n) r_value[j] = v[j-1]; else r_value[j] = INT_MAX; } //cout<<"22222"<<endl; int s,k; cin>>s>>k; k--; for(int u=1;u<=k;u++) { if(l_value[s]<r_value[s]) { int temp = s; s = l_room[s]; r_room[s] = r_room[temp]; r_value[s] = r_value[temp]; } else { int temp = s; s = r_room[s]; l_room[s] = l_room[temp]; l_value[s] = l_value[temp]; } } result.push_back(s); //cout<<"end this"<<endl; } //end of q printf("Case #%d: ",i); for(auto u:result) cout<<u<<" "; cout<<endl; } return 0; }
kickstart 谷歌 D 2020 年 7 月 12 日 13: 00 - 16: 00
标签：rand absolute ror amp googl error Once visitor pre
原文地址：https://www.cnblogs.com/qiezi-online/p/13288684.html
