A. Vladik and Courtesy
题面
At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.
More formally, the guys take turns giving each other one candy more than they received in the previous turn.
This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.
题意
没人轮流减去一个数字，看谁先挂。
代码
#include<bits/stdc++.h>
using namespace std;
int n,m,cnt;
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
cnt=1;
while (cnt )
{
if (cnt%2) n-=cnt;
else m-=cnt;
if (n<0) return 0*puts("Vladik");
if (m<0) return 0*puts("Valera");
cnt++;
}
}
B. Vladik and Complicated Book
题面
Vladik had started reading a complicated book about algorithms containing n pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation P = [p1, p2, ..., pn], where pi denotes the number of page that should be read i-th in turn.
Sometimes Vladik’s mom sorted some subsegment of permutation P from position l to position r inclusive, because she loves the order. For every of such sorting Vladik knows number x — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has px changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.
题意
问子区间排序后第x个值的值会不会改变
找出排列后他应该在哪里，直接判断。
代码
#include<bits/stdc++.h>
using namespace std;
int n,m;
int a[10001];
int b[10001];
int l,r,x;
int cnt;
int cmp(int q,int w)
{
return q<w;
}
int main()
{
ios::sync_with_stdio(false);
cin>>n>>m;
for (int i=1;i<=n;i++) cin>>a[i];
for (int i=1;i<=m;i++)
{
cin>>l>>r>>x;
cnt=0;
for (int o=l;o<=r;o++) if (a[o] < a[x] ) cnt++;
puts( (l+cnt==x) ?"Yes":"No");
}
}
C. Vladik and Memorable Trip
题面
Vladik often travels by trains. He remembered some of his trips especially well and I would like to tell you about one of these trips:
Vladik is at initial train station, and now n people (including Vladik) want to get on the train. They are already lined up in some order, and for each of them the city code ai is known (the code of the city in which they are going to).
Train chief selects some number of disjoint segments of the original sequence of people (covering entire sequence by segments is not necessary). People who are in the same segment will be in the same train carriage. The segments are selected in such way that if at least one person travels to the city x, then all people who are going to city x should be in the same railway carriage. This means that they can’t belong to different segments. Note, that all people who travel to the city x, either go to it and in the same railway carriage, or do not go anywhere at all.
Comfort of a train trip with people on segment from position l to position r is equal to XOR of all distinct codes of cities for people on the segment from position l to position r. XOR operation also known as exclusive OR.
Total comfort of a train trip is equal to sum of comfort for each segment.
Help Vladik to know maximal possible total comfort.
题意
给一个数组，如果你选了一个数，那么就必须选上含这个数的最小区间。
我们可以预处理，得到很多线段，然后我们想办法合并这些线段。
枚举每一个线内，如果有一个区间覆盖了右边的某部分，则合并进来，如果覆盖了左边的某部分，说明之前判断过，如果完全覆盖了，说明也是之前判断过，合并过了。
然后dp，一个线段，d[线段头]转移到d[线段尾部]。
代码
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
int n;
int a[5011];
int in[5011];
int ini[5011];
int last[5011];
ll d[5011];
vector< pair< int , ll > > v[5011];
map<int, pair<int,int> > m;
int main()
{
ios::sync_with_stdio(false);
cin>>n;
for (int i=1;i<=n;i++)
{
cin>>a[i];
last[a[i]]=i;
}
for (int i=1;i<=n;i++) if (!in[a[i]])
{
m[a[i]] = make_pair( i, last[a[i]] );
in[a[i]]=1;
}
for (auto p : m)
{
memset(ini,0,sizeof ini);
auto pi = p.second;
bool flag=0;
ll cos=0;
for (int i=p.second.first;i<=p.second.second;i++)
{
if (m[a[i]].first<p.second.first && m[a[i]].second>p.second.second)
{
flag=1;
break;
}
if (pi.first>m[a[i]].first)
{
flag=1;
break;
}
if (!ini[a[i]])
{
cos^=a[i];
ini[a[i]]=1;
}
pi.second= max(m[a[i]].second,pi.second);
}
if (flag) continue;
v[pi.first].push_back(make_pair(pi.second,cos));
}
int j=0;
for (int i=0;i<=n;i++)
{
for (auto j : v[i])
d[j.first]= max(d[j.first],d[i]+j.second);
/*for (auto j : v[i])
cout<<i<<" "<<j.first<<" "<<j.second<<endl;*/
d[i+1]=max(d[i+1],d[i]);
}
cout<<d[n+1];
}
D. Vladik and Favorite Game
题面
This is an interactive problem.
Vladik has favorite game, in which he plays all his free time.
Game field could be represented as n × m matrix which consists of cells of three types:
- «.» — normal cell, player can visit it.
- «F» — finish cell, player has to finish his way there to win. There is exactly one cell of this type.
- «*» — dangerous cell, if player comes to this cell, he loses.
Initially player is located in the left top cell with coordinates (1, 1).
Player has access to 4 buttons "U", "D", "L", "R", each of them move player up, down, left and right directions respectively.
But it’s not that easy! Sometimes friends play game and change functions of buttons. Function of buttons "L" and "R" could have been swapped, also functions of buttons "U" and "D" could have been swapped. Note that functions of buttons can be changed only at the beginning of the game.
Help Vladik win the game!
题意
交互题目，感觉比C要简单很多
先bfs一个可行路径出来，然后就跟着走，如果走了以后没效果（原地不动），那就反着走
代码
#include<bits/stdc++.h>
using namespace std;
int n,m;
string d;
char a[110][110];
int vis[110][110];
string moves[]={"L","R","U","D"};
int dx[]={0,0,-1,1};
int dy[]={-1,1,0,0};
string ans;
string newans;
queue<pair<int,int> > q;
queue<string> qs;
int main()
{
cin>>n>>m;
for (int i=1;i<=n;i++)
{
cin>>d;
for (int o=1;o<=m;o++) a[i][o]=d[o-1];
}
q.push(make_pair(1,1));
qs.push("");
while (!q.empty())
{
int xx=q.front().first;
int yy=q.front().second;
q.pop();
string now=qs.front();qs.pop();
for (int k=0;k<4;k++)
{
if (xx+dx[k]>n || xx+dx[k]<1) continue;
if (yy+dy[k]>m || yy+dy[k]<1) continue;
if (vis[xx+dx[k]][yy+dy[k]]) continue;
if (a[xx+dx[k]][yy+dy[k]]!='*')
{
vis[xx+dx[k]][yy+dy[k]]=1;
q.push(make_pair(xx+dx[k],yy+dy[k]));
qs.push(now+moves[k]);
if (a[xx+dx[k]][yy+dy[k]]=='F')
{
ans=now+moves[k];
}
}
if (ans!="") break;
}
if (ans!="") break;
}
if (ans=="") return 0;
//cout<<ans<<endl;
int nowx=1;
int nowy=1;
int x,y;
bool flag1=0;
bool flag2=0;
for (int i=0;i<ans.size();i++)
{
cout<<ans[i]<<endl;
fflush(stdout);
cin>>x>>y;
if (x==nowx && y==nowy && (ans[i]=='L' || ans[i]=='R') && !flag1)
{
flag1=1;
newans.clear();
for (int o=0;o<ans.size();o++)
{
if (ans[o] == 'R') newans.push_back('L');
else if (ans[o] == 'L') newans.push_back('R');
else newans.push_back(ans[o]);
}
//cout<<newans<<endl;
ans=newans;
i--;
}
else if (x==nowx && y==nowy && (ans[i]=='U' || ans[i]=='D') && !flag2)
{
flag2=1;
newans.clear();
for (int o=0;o<ans.size();o++)
{
if (ans[o] == 'U') newans.push_back('D');
else if (ans[o] == 'D') newans.push_back('U');
else newans.push_back(ans[o]);
}
//cout<<newans<<endl;
ans=newans;
i--;
}
nowx=x;
nowy=y;
}
}
E. Vladik and Entertaining Flags
题面
In his spare time Vladik estimates beauty of the flags.
Every flag could be represented as the matrix n × m which consists of positive integers.
Let's define the beauty of the flag as number of components in its matrix. We call component a set of cells with same numbers and between any pair of cells from that set there exists a path through adjacent cells from same component. Here is the example of the partitioning some flag matrix into components:
But this time he decided to change something in the process. Now he wants to estimate not the entire flag, but some segment. Segment of flag can be described as a submatrix of the flag matrix with opposite corners at (1, l) and (n, r), where conditions 1 ≤ l ≤ r ≤ m are satisfied.
Help Vladik to calculate the beauty for some segments of the given flag.
题意
给你一个矩形，和几个询问，问从l，r之间的列形成的矩形，的消消乐块数是多少
我很诡异的解释qwq
这题跟之前做到的atcoder题目很类似，下次一起补完
代码
//
比赛总结
这场本来感觉是要掉rating掉很多的，原因有很多方面，一个是C看错题目了，导致没写出来。
然后D搜索的时候，想着搜一条最短路，导致超时，很不应该。
E题感觉算是种套路，atcoder里面的感觉更加凶残一点。
话说我Atcoder就没做过2题以上。。