Codeforces Round #516 (Div. 2, by Moscow Team Olympiad)

Posted 2021-02-13 fighting-sh

Codeforces Round #516 (Div. 2, by Moscow Team Olympiad)

https://codeforces.com/contest/1064

A

 1 #include<bits/stdc++.h>
 2 #define pb push_back
 3 using namespace std;
 4 
 5 bool Check(int a,int b,int c){
 6     if(a+b>c&&b+c>a&&a+c>b) return true;
 7     return false;
 8 }
 9 
10 int main(){
11     int a,b,c;
12     cin>>a>>b>>c;
13     if(Check(a,b,c)){
14         cout<<0<<endl;
15     }
16     else{
17         cout<<min(abs(a+b-c),min(abs(b+c-a),abs(a+c-b)))+1<<endl;
18 
19     }
20 }

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B

打表找规律

 1 #include<bits/stdc++.h>
 2 #define pb push_back
 3 typedef long long ll;
 4 using namespace std;
 5 
 6 
 7 int main(){
 8     int t;
 9   /*  for(int i=0;i<=100;i++){
10         int co=0;
11         for(int j=0;j<=i;j++){
12             if((i-j-(i^j))==0){
13                 co++;
14             }
15         }
16         cout<<i<<" "<<co<<endl;
17     }*/
18 
19     cin>>t;
20     int n;
21     while(t--){
22         cin>>n;
23         int num=__builtin_popcount(n);
24         ll ans=pow(2,num);
25         cout<<ans<<endl;
26     }
27 }

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C

题意：重新排序字符串，使它的子字符串是回文串的数量最多

思路：把相同的字母排在一起即可

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define eb emplace_back
 8 #define maxn 300005
 9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<long long,int>pli;
15 typedef pair<int,char> pic;
16 typedef pair<pair<int,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long mod=1e9+7;
19 using namespace std;
20 
21 int n;
22 string str;
23 map<char,int>mp;
24 map<char,int>::iterator it;
25 string ans;
26 vector<char>ve;
27 int main(){
28     cin>>n>>str;
29     for(int i=0;i<n;i++){
30         ve.pb(str[i]);
31     }
32     sort(ve.begin(),ve.end());
33     for(int i=0;i<ve.size();i++) cout<<ve[i];
34 
35 }

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D

题意：一个人可以上下左右移动，但是左右移动有次数限制，求他活动范围最多是多少个格子

思路：bfs搜索，用book数组标记剩余的左右移动的次数

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define eb emplace_back
 8 #define maxn 300005
 9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<long long,int>pli;
15 typedef pair<int,char> pic;
16 typedef pair<pair<int,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long mod=1e9+7;
19 using namespace std;
20 
21 int n,m;
22 char str[2005][2005];
23 struct Num{
24     int L,R;
25 }book[2005][2005];
26 int r,c,L,R;
27 struct sair{
28     int x,y,L,R;
29 };
30 int dir[4][2]={0,-1,-1,0,0,1,1,0};//R,D,L,U
31 
32 void bfs(){
33     queue<sair>Q;
34     sair s,e;
35     s.x=r,s.y=c,s.L=L,s.R=R;
36     Q.push(s);
37     book[s.x][s.y].L=L;
38     book[s.x][s.y].R=R;
39     while(!Q.empty()){
40         s=Q.front();
41         Q.pop();
42         for(int i=0;i<4;i++){
43             e.x=s.x+dir[i][0];
44             e.y=s.y+dir[i][1];
45             if(e.x>=0&&e.x<n&&e.y>=0&&e.y<m&&str[e.x][e.y]!=‘*‘){
46                 e.L=s.L;
47                 e.R=s.R;
48                 if(i==2){
49                     e.R--;
50                     if(e.R<0) continue;
51                 }
52                 else if(i==0){
53                     e.L--;
54                     if(e.L<0) continue;
55                 }
56                 if(book[e.x][e.y].L<e.L||book[e.x][e.y].R<e.R){
57                     book[e.x][e.y].L=max(book[e.x][e.y].L,e.L);
58                     book[e.x][e.y].R=max(book[e.x][e.y].R,e.R);
59                     Q.push(e);
60                 }
61 
62             }
63         }
64     }
65 }
66 
67 int main(){
68     std::ios::sync_with_stdio(false);
69     cin>>n>>m;
70     cin>>r>>c;
71     cin>>L>>R;
72     for(int i=0;i<=2000;i++){
73         for(int j=0;j<=2000;j++){
74             book[i][j].L=book[i][j].R=-1;
75         }
76     }
77     for(int i=0;i<n;i++){
78         cin>>str[i];
79     }
80     r--,c--;
81     bfs();
82     int ans=0;
83     for(int i=0;i<n;i++){
84         for(int j=0;j<m;j++){
85             if(book[i][j].L!=-1||book[i][j].R!=-1){
86                 ans++;
87             }
88         }
89     }
90     cout<<ans<<endl;
91 }

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E

交互题

题意：输出n个点，每输出一个点会告诉你这点是黑色还是白色 ，最后输出一条线段，使黑白色的点分别在线段的两边

思路：二分，当点的颜色与第一个点相同时，往右走，否则往左走，这样就把不同颜色的点分成两边。因为n为30，范围为1e9，所以可以二分30次

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define eb emplace_back
 8 #define maxn 300005
 9 #define eps 1e-8
10 #define pi acos(-1.0)
11 #define rep(k,i,j) for(int k=i;k<j;k++)
12 typedef long long ll;
13 typedef pair<int,int> pii;
14 typedef pair<long long,int>pli;
15 typedef pair<int,char> pic;
16 typedef pair<pair<int,string>,pii> ppp;
17 typedef unsigned long long ull;
18 const long long mod=1e9+7;
19 using namespace std;
20 
21 
22 int main(){
23     std::ios::sync_with_stdio(false);
24     int n;
25     cin>>n;
26     string str,s;
27     int L=0,R=1e9;
28     int ans=0;
29     cout<<ans<<" "<<1<<endl;
30     cin>>s;
31     for(int i=1;i<n;i++){
32         ans=L+R>>1;
33         cout<<ans<<" "<<1<<endl;
34         cin>>str;
35         if(str==s){
36             L=ans;
37         }
38         else{
39             R=ans;
40         }
41     }
42     cout<<L<<" "<<2<<" "<<R<<" "<<0<<endl;
43 }

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